Classical Geometry: Euclidean, Transformational, Inversive, and Projective
I. E. Leonard, J. E. Lewis, A. C. F. Liu, G. W. TokarskyFeatures the classical themes of geometry with plentiful applications in mathematics, education, engineering, and science
Accessible and readerfriendly, Classical Geometry: Euclidean, Transformational, Inversive, and Projective introduces readers to a valuable discipline that is crucial to understanding bothspatial relationships and logical reasoning. Focusing on the development of geometric intuitionwhile avoiding the axiomatic method, a problem solving approach is encouraged throughout.
The book is strategically divided into three sections: Part One focuses on Euclidean geometry, which provides the foundation for the rest of the material covered throughout; Part Two discusses Euclidean transformations of the plane, as well as groups and their use in studying transformations; and Part Three covers inversive and projective geometry as natural extensions of Euclidean geometry. In addition to featuring realworld applications throughout, Classical Geometry: Euclidean, Transformational, Inversive, and Projective includes:
 Multiple entertaining and elegant geometry problems at the end of each section for every level of study
 Fully worked examples with exercises to facilitate comprehension and retention
 Unique topical coverage, such as the theorems of Ceva and Menalaus and their applications
 An approach that prepares readers for the art of logical reasoning, modeling, and proofs
The book is an excellent textbook for courses in introductory geometry, elementary geometry, modern geometry, and history of mathematics at the undergraduate level for mathematics majors, as well as for engineering and secondary education majors. The book is also ideal for anyone who would like to learn the various applications of elementary geometry.
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About the pagination of this eBook Due to the unique page numbering scheme of this book, the electronic pagination of the eBook does not match the pagination of the printed version. To navigate the text, please use the electronic Table of Contents that appears alongside the eBook or the Search function. For citation purposes, use the page numbers that appear in the text. CLASSICAL GEOMETRY CLASSICAL GEOMETRY Euclidean, Transformational, Inversive, and Projective I. E. Leonard, J. E. Lewis, A. C. F. Liu, G. W. Tokarsky Department of Mathematical and Statistical Sciences University of Alberta WILEY Copyright© 2014 by John Wiley & Sons, Inc. All rights reserved. Published by John Wiley & Sons, Inc., Hoboken, New Jersey. Published simultaneously in Canada. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission ofthe Publisher, or authorization through payment of the appropriate percopy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, (978) 7508400, fax (978) 7504470, or on the web at www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 7486011, fax (201) 7486008, or online at http://www.wiley.com/go/permission. Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts in preparing this book, they make no representation or warranties with respect to the accuracy or completeness of the contents of this book and specifically disclaim any implied warranties of merchantability or fitness for a particular purpose. No warranty may be created or extended by sales representatives or written sales materials. The advice and strate; gies contained herein may not be suitable for your situation. You should consult with a professional where appropriate. Neither the publisher nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. For general information on our other products and services please contact our Customer Care Department within the United States at (800) 7622974, outside the United States at (317) 5723993 or fax (317) 5724002. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print, however, may not be available in electronic formats. For more information about Wiley products, visit our web site at www.wiley.com. Library of Congress CataloginginPublication Data: Leonard, I. Ed., 1938 author. Classical geometry : Euclidean, transformational, inversive, and projective I I. E. Leonard, Department of Mathematical and Statistical Sciences, University of Alberta, Edmonton, Canada, J. E. Lewis, Department of Mathematical and Statistical Sciences, University of Alberta, Edmonton, Canada, A. C. F. Liu, Department of Mathematical and Statistical Sciences, University of Alberta, Edmonton, Canada, G. W. Tokarsky, Department of Mathematical and Statistical Sciences, University of Alberta, Edmonton, Canada. pages em Includes bibliographical references and index. ISBN 9781118679197 (hardback) 1. Geometry. I. Lewis, J. E. (James Edward) author. II. Liu, A. C. F. (Andrew ChiangFung) author. III. Tokarsky, G. W., author. IV. Title. QA445.L46 2014 516dc23 2013042035 Printed in the United States of America. 10 9 8 7 6 5 4 3 2 1 CONTENTS xi Preface PART I 1 EUCLIDEAN GEOMETRY Congruency 3 1.1 1.2 1.3 3 1.4 1.5 1.6 1.7 1.8 Introduction Congruent Figures Parallel Lines 1.3.1 Angles in a Triangle Thales' Theorem 1.3.2 1.3.3 Quadrilaterals More About Congruency Perpendiculars and Angle Bisectors Construction Problems 1.6.1 The Method of Loci Solutions to Selected Exercises Problems 6 12 13 14 17 21 24 28 31 33 38 v vi CONTENTS 2 Concurrency 41 2.1 2.2 2.3 2.4 2.5 2.6 2.7 41 43 46 48 50 54 56 3 4 Similarity 59 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 59 60 64 67 75 82 87 90 Similar Triangles Parallel Lines and Similarity Other Conditions Implying Similarity Examples Construction Problems The Power of a Point Solutions to the Exercises Problems Theorems of Ceva and Menelaus 4.1 4.2 4.3 4.4 4.5 4.6 4.7 5 Perpendicular Bisectors Angle Bisectors Altitudes Medians Construction Problems Solutions to the Exercises Problems Directed Distances, Directed Ratios The Theorems Applications of Ceva's Theorem Applications of Menelaus' Theorem Proofs of the Theorems Extended Versions of the Theorems 4.6.1 Ceva's Theorem in the Extended Plane 4.6.2 Menelaus' Theorem in the Extended Plane Problems 5.2 95 97 99 103 115 125 127 129 131 133 Area 5.1 95 Basic Properties 5.1.1 Areas of Polygons 5.1.2 Finding the Area of Polygons 5.1.3 Areas of Other Shapes Applications of the Basic Properties 133 134 138 139 140 CONTENTS 6 5.3 Other Formulae for the Area of a Triangle 5.5 Problems 159 6.1 6.2 6.3 159 161 166 168 169 171 178 185 188 193 197 200 201 6.7 6.8 6.9 6.10 The Three Problems of Antiquity Constructing Segments of Specific Lengths Construction of Regular Polygons 6.3.1 Construction of the Regular Pentagon 6.3.2 Construction of Other Regular Polygons Miquel's Theorem Morley's Theorem The NinePoint Circle 6.6.1 Special Cases The SteinerLehmus Theorem The Circle of Apollonius Solutions to the Exercises Problems PART II TRANSFORMATIONAL GEOMETRY The Euclidean Transformations or lsometries 207 7.1 7.2 207 211 213 217 227 7.3 7.4 8 147 5.4 Solutions 153 153 Miscellaneous Topics 6.4 6.5 6.6 7 vii Rotations, Reflections, and Translations Mappings and Transformations 7.2.1 Isometries Using Rotations, Reflections, and Translations Problems The Algebra of lsometries 235 8.1 8.2 235 240 241 245 250 8.3 8.4 Basic Algebraic Properties Groups of Isometries 8.2.1 Direct and Opposite Isometries The Product of Reflections Problems viii 9 10 CONTENTS The Product of Direct lsometries 255 9.1 9.2 9.3 9.4 9.5 9.6 255 257 258 259 262 265 Symmetry and Groups 271 10.1 271 275 279 283 10.2 10.3 11 More About Groups 10.1.1 Cyclic and Dihedral Groups Leonardo's Theorem Problems Homotheties 289 11.1 11.2 289 290 293 295 300 304 306 11.3 11.4 11.5 11.6 12 Angles Fixed Points The Product of Two Translations The Product of a Translation and a Rotation The Product of Two Rotations Problems The Pantograph Some Basic Properties 11.2.1 Circles Construction Problems Using Homotheties in Proofs Dilatation Problems Tessellations 313 12.1 12.2 12.3 12.4 12.5 313 314 319 325 332 Tilings Monohedral Tilings Tiling with Regular Polygons Platonic and Archimedean Tilings Problems PART Ill 13 INVERSIVE AND PROJECTIVE GEOMETRIES Introduction to Inversive Geometry 339 13.1 13.2 13.3 13.4 13.5 339 345 353 362 371 Inversion in the Euclidean Plane The Effect of Inversion on Euclidean Properties Orthogonal Circles CompassOnly Constructions Problems CONTENTS 14 15 16 ix Reciprocation and the Extended Plane 375 14.1 14.2 14.3 14.4 14.5 375 385 396 402 409 Harmonic Conjugates The Projective Plane and Reciprocation Conjugate Points and Lines Conics Problems Cross Ratios 411 15.1 15.2 15.3 Cross Ratios Applications of Cross Ratios Problems 411 422 431 Introduction to Projective Geometry 435 16.1 16.2 16.3 16.4 16.5 16.6 16.7 435 445 450 450 453 457 463 Straightedge Constructions Perspectivities and Projectivities Line Perspectivities and Line Projectivities Projective Geometry and Fixed Points Projecting a Line to Infinity The Apollonian Definition of a Conic Problems Bibliography 466 Index 471 PREFACE It is sometimes said that geometry should be studied because it is a useful and valuable discipline, but in fact many people study it simply because geometry is a very enjoyable subject. It is filled with problems at every level that are entertaining and elegant, and this enjoyment is what we have attempted to bring to this textbook. This text is based on class notes that we developed for a threesemester sequence of undergraduate geometry courses that has been taught at the University of Alberta for many years. It is appropriate for students from all disciplines who have previously studied high school algebra, geometry, and trigonometry. When we first started teaching these courses, our main problem was finding a suitable method for teaching geometry to university students who have had minimal experience with geometry in high school. We experimented with material from high school texts but found it was not challenging enough. We also tried an axiomatic approach, but students often showed little enthusiasm for proving theorems, particularly since the early theorems seemed almost as selfevident as the axioms. We found the most success by starting early with problem solving, and this is the approach we have incorporated throughout the book. xi Xii PREFACE The geometry in this text is synthetic rather than Cartesian or coordinate geometry. We remain close to classical themes in order to encourage the development of geometric intuition, and for the most part we avoid abstract algebra although we do demonstrate its use in the sections on transformational geometry. Part I is about Euclidean geometry; that is, the study of the properties of points and lines that are invariant under isometries and similarities. As well as many of the usual topics, it includes material that many students will not have seen, for example, the theorems of Ceva and Menelaus and their applications. Part I is the basis for Parts II and III. Part II discusses the properties of Euclidean transformations or isometries of the plane (translations, reflections, and rotations and their compositions). It also introduces groups and their use in studying transformations. Part III introduces inversive and projective geometry. These subjects are presented as natural extensions of Euclidean geometry, with no abstract algebra involved. We would like to acknowledge our late colleagues George Cree and Murray Klarnkin, without whose inspiration and encouragement over the years this project would not have been possible. Finally, we would like to thank our families for their patience and understanding in the preparation of the textbook. In particular, I. E. Leonard would like to thank Sarah for proofreading the manuscript numerous times. ED, TED, ANDY, AND GEORGE Edmonton, Alberta, Canada January, 2014 PART I EUCLIDEAN GEOMETRY CHAPTER 1 CONGRUENCY 1.1 Introduction Assumed Knowledge This text assumes a bit of knowledge on the part of the reader. For example, it assumes that you know that the sum of the angles of a triangle in the plane is 180° (x + y + z = 180° in the figure below), and that in a right triangle with hypotenuse c and sides a and b, the Pythagorean relation holds: c2 = a 2 + b2 . a Classical Geometry: Euclidean, Transformational, Inversive and Projective, First Edition. By I. E. Leonard, J. E. Lewis, A. C. F. Liu, G. W. Tokarsky Copyright© 2014 John Wiley & Sons, Inc. 3 4 CONGRUENCY We use the word line to mean straight line, and we assume that you know that two lines either do not intersect, intersect at exactly one point, or completely coincide. Two lines that do not intersect are said to be parallel. We also assume certain knowledge about parallel lines, namely, that you have seen some form of the parallel axiom: Given a line l and a point P in the plane, there is exactly one line through P parallel to l. p ·l The preceding version of the parallel axiom is often called Playfair's Axiom. You may even know something equivalent to it that is close to the original version of the parallel postulate: Given two lines land m, and a third line t cutting both land m and forming angles ¢>and() on the same side oft, if¢>+() < 180°, then land m meet at a point on the same side oft as the angles. m p )o l The subject of this part of the text is Euclidean geometry, and the abovementioned parallel postulate characterizes Euclidean geometry. Although the postulate may seem to be obvious, there are perfectly good geometries in which it does not hold. INTRODUCTION 5 We also assume that you know certain facts about areas. A parallelogram is a quadrilateral (figure with four sides) such that the opposite sides are parallel. The area of a parallelogram with base b and height h is b · h, and the area of a triangle with baseband height his b · h/ 2. b Notation and Terminology Throughout this text, we use uppercase Latin letters to denote points and lowercase Latin letters to denote lines and rays. Given two points A and B, there is one and ; only one line through A and B. A ray is a halfline, and the notation AB denotes the ray starting at A and passing through B. It consists of the points A and B, all points between A and B, and all points X on the line such that B is between A and X. ; ; Given rays AB and AC, we denote by L.BAC the angle formed by the two rays (the shaded region in the following figure). When no confusion can arise, we sometimes use L.A instead of L.BAC. We also use lowercase letters, either Greek or Latin, to denote angles. ~· · ·· <P\ J <P = 360 e c When two rays form an angle other than 180°, there are actually two angles to talk about: the smaller angle (sometimes called the interior angle) and the larger angle (called the reflex angle). When we refer to L.BAC, we always mean the nonreftex angle. Note. The angles that we are talking about here are undirected angles; that is, they do not have negative values, and so can range in magnitude from 0° to 360°. Some people prefer to use m(L.A) for the measure of the angle A; however, we will use the same notation for both the angle and the measure of the angle. 6 CONGRUENCY When we refer to a quadrilateral as ABCD we mean one whose edges are AB, BC, CD, and DA. Thus, the quadrilateral ABC D and the quadrilateral ABDC are quite different. D b~ C D C There are three classifications of quadrilaterals: convex, simple, and nonsimple, as shown in the following diagram. nonsimple quadrilateral convex quadrilateral simple quadrilaterals quadrilaterals 1.2 Congruent Figures Two figures that have exactly the same shape and exactly the same size are said to be congruent. More explicitly: 1. Two angles are congruent if they have the same measure. 2. Two line segments are congruent if they are the same length. 3. Two circles are congruent if they have the same radius. 4. Two triangles are congruent if corresponding sides and angles are the same size. 5. All rays are congruent. 6. All lines are congruent. Theorem 1.2.1. Vertically opposite angles are congruent. Proof. We want to show that a have a +c = b. We = 180 and b + c = 180, and it follows from this that a = b. D CONGRUENT FIGURES 7 Notation. The symbol = denotes congruence. We use the notation 6ABC to denote a triangle with vertices A, B, and C, and we use C(P, r) to denote a circle with center P and radius r. Thus, C(P, r) =C(Q, s) if and only if r = s. We will be mostly concerned with the notion of congruent triangles, and we mention that in the definition, 6ABC 6DEF if and only if the following six conditions hold: = LA=LD LB=LE LC=:LF AB=DE BC=:EF AC Note that the two statements 6ABC same! =DF. =6DEF and 6ABC =6EF Dare not the The Basic Congruency Conditions According to the definition of congruency, two triangles are congruent if and only if six different parts of one are congruent to the six corresponding parts of the other. Do we really need to check all six items? The answer is no. If you give three straight sticks to one person and three identical sticks to another and ask both to constuct a triangle with the sticks as the sides, you would expect the two triangles to be exactly the same. In other words, you would expect that it is possible to verify congruency by checking that the three corresponding sides are congruent. Indeed this is the case, and, in fact, there are several ways to verify congruency without checking all six conditions. The three congruency conditions that are used most often are the SideAngleSide (SAS) condition, the SideSideSide (SSS) condition, and the AngleSideAngle (ASA) condition. 8 CONGRUENCY Axiom 1.2.2. (SAS Congruency) Two triangles are congruent if two sides and the included angle of one are congruent to two sides and the included angle of the other. Theorem 1.2.3. (SSS Congruency) Two triangles are congruent if the three sides of one are congruent to the corresponding three sides of the other. Theorem 1.2.4. (ASA Congruency) Two triangles are congruent if two angles and the included side of one are congruent to two angles and the included side of the other. You will note that the SAS condition is an axiom, and the other two are stated as theorems. We will not prove the theorems but will freely use all three conditions. Any one of the three conditions could be used as an axiom with the other two then derived as theorems. In case you are wondering why the SAS condition is preferred as the basic axiom rather than the SSS condition, it is because it is always possible to construct a triangle given two sides and the included angle, whereas it is not always possible to construct a triangle given three sides (consider sides of length 3, 1, and 1). Axiom 1.2.5. (The Triangle Inequality) The sum of the lengths of two sides of a triangle is always greater than the length of the remaining side. The congruency conditions are useful because they allow us to conclude that certain parts of two triangles are congruent by determining that certain other parts are congruent. Here is how congruency may be used to prove two wellknown theorems about isosceles triangles. (An isosceles triangle is one that has two equal sides.) Theorem 1.2.6. (The Isosceles Triangle Theorem) In an isosceles triangle, the angles opposite the equal sides are equal. CONGRUENT FIGURES Proof. Let us suppose that the triangle is ABC with AB 9 = AC. (same triangle, recopied for reference purposes) In !:::,.ABC and 6.AC B we have AB = AC, LBAC=LCAB, AC=AB, so 6.ABC = 6.AC B by SAS. Since the triangles are congruent, it follows that all corresponding parts are congruent, so LB of !:::,.ABC must be congruent to LC of 6.AC B. D Theorem 1.2.7. (Converse of the Isosceles Triangle Theorem) /fin !:::,.ABC we have LB = LC, then AB = AC. Proof. In !:::,.ABC and 6.ACB we have LABC = LACB, BC=CB, LACB = LABC, so 6.ABC = 6.AC B by ASA. Since 6.ABC = 6.ACB it follows that AB = AC. D Perhaps now is a good time to explain what the converse of a statement is. Many statements in mathematics have the form lfP, then Q, where P and Q are assertions of some sort. 10 CONGRUENCY For example: If ABC D is a square, then angles A, B, C, and D are all right angles. Here, Pis the assertion "ABCD is a square," and Q is the assertion "angles A, B, C, and Dare all right angles." The converse of the statement "If P, then Q" is the statement If Q, then P. Thus, the converse of the statement "If ABCD is a square, then angles A, B, C, and D are all right angles" is the statement If angles A, B, C, and Dare all right angles, then ABCD is a square. A common error in mathematics is to confuse a statement with its converse. Given a statement and its converse, if one of them is true, it does not automatically follow that the other is also true. Exercise 1.2.8. For each of the following statements, state the converse and determine whether it is true or false. I. Given triangle ABC, if LABC is a right angle, then AB 2 + BC 2 = AC2 . 2. If ABC D is a parallelogram, then AB = CD and AD = BC. 3. If ABC D is a convex quadrilateral, then ABCD is a rectangle. 4. Given quadrilateral ABCD, if AC# BD, then ABCD is not a rectangle. Solutions to the exercises are given at the end of the chapter. The Isosceles Triangle Theorem and its converse raise questions about how sides are related to unequal angles, and there are useful theorems for this case. Theorem 1.2.9. (The AngleSide Inequality) In 6ABC, if LABC > LAC B, then AC > AB. Proof. Draw a ray BX so that LCBX = LBCA with X to the same side of BC as A, as in the figure on the following page. CONGRUENT FIGURES 11 A Since LABC > LCBX, the point X is interior to LABC and so BX will cut side AC at a point D. Then we have DB=DC by the converse to the Isosceles Triangle Theorem. By the Triangle Inequality, we have AB <AD+DB, and combining these gives us AB < AD+ DC = AC, which is what we wanted to prove. D The converse of the AngleSide Inequality is also true. Note that the proof of the converse uses the statement of the original theorem. This is something that frequently occurs when proving that the converse is true. Theorem 1.2.10. In 6ABC, if AC > AB, then LABC > LAC B. Proof. There are three possible cases to consider: (1) LABC = LACB. (2) LABC < LAC B. (3) LABC > LACB. If case (1) arises, then AC =ABby the converse to the Isosceles Triangle Theorem, so case (1) cannot in fact arise. If case (2) arises, then AC < ABby the AngleSide Inequality, so (2) cannot arise. The only possibility is therefore case (3). D CONGRUENCY 12 The preceding examples, as well as showing how congruency is used, are facts that are themselves very useful. They can be summarized very succinctly: in a triangle, Equal angles are opposite equal sides. The larger angle is opposite the larger side. 1.3 Parallel Lines Two lines in the plane are parallel if (a) they do not intersect or (b) they are the same line. Note that (b) means that a line is parallel to itself. Notation. We use l II m to denote that the lines l and m are parallel and sometimes use l }f m to denote that they are not parallel. If l and m are not parallel, they meet at precisely one point in the plane. When a transversal crosses two other lines, various pairs of angles are endowed with special names: adjacent angles opposite or alternate angles opposite or alternate exterior angles corresponding angles The proofs of the next two theorems are omitted; however, we mention that the proof of Theorem 1.3.2 requires the parallel postulate, but the proof of Theorem 1.3.1 does not. PARALLEL LINES 13 Theorem 1.3.1. If a transversal cuts two lines and any one of the following four conditions holds, then the lines are parallel: (1) adjacent angles total180°, (2) alternate angles are equal, (3) alternate exterior angles are equal, (4) corresponding angles are equal. Theorem 1.3.2. If a transversal cuts two parallel lines, then all four statements of Theorem 1.3.1 hold. Remark. Theorem 1.3.1 can be proved using the External Angle Inequality, which is described below. The proof of the inequality itself ultimately depends on Theorem 1.3.1, but this would mean that we are using circular reasoning, which is not permitted. However, there is a proof of the External Angle Inequality which does not in any way depend upon Theorem 1.3.1, and so it is possible to avoid circular reasoning. 1.3.1 Angles in a Triangle The parallel postulate is what distinguishes Euclidean geometry from other geometries, and as we see now, it is also what guarantees that the sum of the angles in a triangle is 180°. Theorem 1.3.3. The sum of the angles of a triangle is 180°. Proof. Given triangle ABC, draw the line XY through A parallel to BC, as shown. Consider AB as a transversal for the parallel lines XY and BC, then x = u and similarly y = v. Consequently, X+ y + W = U + V + W = 180, which is what we wanted to prove. D 14 CONGRUENCY Given triangle ABC, extend the side BC beyond C to X. The angle ACX is called an exterior angle of 6.ABC. Theorem 1.3.4. (The Exterior Angle Theorem) An exterior angle of a triangle is equal to the sum of the opposite interior angles. A Proof. In the diagram above, we have y +z = 180 = y + x + w, so z = x + w. 0 The Exterior Angle Theorem has a useful corollary: Corollary 1.3.5. (The Exterior Angle Inequality) An exterior angle of a triangle is greater than either of the opposite interior angles. Note. The proof of the Exterior Angle Inequality given above ultimately depends on the fact that the sum of the angles of a triangle is 180°, which turns out to be equivalent to the parallel postulate. It is possible to prove the Exterior Angle Inequality without using any facts that follow from the parallel postulate, but we will omit that proof here. 1.3.2 Thales' Theorem One of the most useful theorems about circles is credited to Thales, who is reported to have sacrificed two oxen after discovering the proof. (In truth, versions of the theorem were known to the Babylonians some one thousand years earlier.) Theorem 1.3.6. (Thales' Theorem) An angle inscribed in a circle is half the angle measure of the intercepted arc. PARALLEL LINES 15 In the diagram, o: is the measure of the inscribed angle, the arc CD is the intercepted arc, and {3 is the angle measure of the intercepted arc. The following diagrams illustrate Thales' Theorem. Proof. As the figures above indicate, there are several separate cases to consider. We will prove the first case and leave the others as exercises. Referring to the diagram below, we have o: = v +w and {3 = f.L + ry. But v =wand J.L = 'TJ (isosceles triangles). Consequently, L.BOC = o: + {3 = 2w + 2ry = 2L.BAC, and the theorem follows. 0 16 CONGRUENCY Thales' Theorem has several useful corollaries. Corollary 1.3.7. In a given circle: ( 1) All inscribed angles that intercept the same arc are equal in size. (2) All inscribed angles that intercept congruent arcs are equal in size. (3) The angle in a semicircle is a right angle. The converse of Thales' Theorem is also very useful. Theorem 1.3.8. (Converse ofThales' Theorem) Let H be a halfplane determined by a line PQ. The set of points in H that form a constant angle (3 with P and Q is an arc of a circle passing through P and Q. Furthermore, every point ofH inside the circle makes a larger angle with P and Q and every point ofH outside the circle makes a smaller angle with P and Q. z Proof. Let S be a point such that L.P SQ = (3 and let C be the circumcircle of !:::,SPQ. In the halfplane H, all points X on C intercept the same arc of C, so by Thales' Theorem, all angles PXQ have measure (3 . PARALLEL LINES 17 From the Exterior Angle Inequality, in the figure on the previous page we have a> fJ > 'Y· Asaconsequence,everypointZof1toutsideCmusthaveLPZQ < {J, and every pointY of 1t inside C must have LPYQ > {J, and this completes the proof. 0 Exercise 1.3.9. Calculate the size of() in the following figure. 1.3.3 Quadrilaterals The following theorem uses the fact that a simple quadrilateral always has at least one diagonal that is interior to the quadrilateral. Theorem 1.3.10. The sum of the interior angles of a simple quadrilateral is 360°. B B Proof. Let the quadrilateral have vertices A, B, C, and D, with AC being an internal diagonal. Referring to the diagram, we have LA+ LB + LC + LD =(¢+a)+ {J + ("! + ()) + 8 =(a+ fJ + 'Y) + (() + 8 + ¢) = 180° = 360°. + 180° 0 18 CONGRUENCY Note. This theorem is false if the quadrilateral is not simple, in which case the sum of the interior angles is less than 360°. Cyclic Quadrilaterals A quadrilateral that is inscribed in a circle is called a cyclic quadrilateral or, equivalently, a concyclic quadrilateral. The circle is called the circumcircle of the quadrilateral. Theorem 1.3.11. Let ABC D be a simple cyclic quadrilateral. Then: (1) The opposite angles sum to 180°. (2) Each exterior angle is congruent to the opposite interior angle. a+P = 180 a=P Theorem 1.3.12. Let ABCD be a simple quadrilateral. If the opposite angles sum to 180°, then ABCD is a cyclic quadrilateral. We leave the proofs of Theorem 1.3.11 and Theorem 1.3.12 as exercises and give a similar result for nonsimple quadrilaterals. Example 1.3.13. (Cyclic Nonsimple Quadrilaterals) A nonsimple quadrilateral can be inscribed in a circle ifand only ifopposite angles are equal. For example, in the figure on the following page, the nonsimple quadrilateral ABC D can be inscribed in a circle if and only if LA = LC and LB = LD. PARALLEL LINES 19 D Solution. Suppose first that the quadrilateral ABCD is cyclic. Then LA = LC since they are both subtended by the chord BD, while LB = LD since they are both subtended by the chord AC. Conversely, suppose that LA= LC and LB circumcircle of 6ABC. = LD, and let the circle below be the D If the quadrilateral ABCD is not cyclic, then the point D does not lie on this circumcircle. Assume that it lies outside the circle and let D' be the point where the line segment AD hits the circle. Since ABCD' is a cyclic quadrilateral, then from the first part of the proof, LB = LD' and therefore LD = LD', which contradicts the External Angle Inequality in 6CD' D. Thus, if LA= LC and LB = LD, then quadrilateral ABCD is cyclic. 0 20 CONGRUENCY Exercise 1.3.14. Show that a quadrilateral has an inscribed circle (that is, a circle tangent to each of its sides) if and only if the sums of the lengths of the two pairs of opposite sides are equal. For example, the quadrilateral ABCD has an inscribed circle if and only if AB + CD = AD + BC. D c The following example is a result named after Robert Simson (16871768), whose Elements of Euclid was a standard textbook published in 24 editions from 1756 until 1834 and upon which many modem English versions of Euclid are based. However, in their book Geometry Revisited, Coxeter and Greitzer report that the result attributed to Simson was actually discovered later, in 1797, by William Wallace. Example 1.3.15. (Simson's Theorem) Given 6.ABC inscribed in a circle and a point P on its circumference, the perpendiculars dropped from P meet the sides of the triangle in three collinear points. The line is called the Simson line corresponding to P. Solution. We will prove Simson's Theorem by showing that LP EF (which means that the rays EF and ED coincide). LPED As well as the cyclic quadrilateral PACB, there are two other cyclic quadrilaterals, namely PEAF and PECD, which are reproduced in the figure on the following page. (These are cyclic because in each case two of the opposite angles sum to 180°). MORE ABOUT CONGRUENCY 21 By Thales' Theorem applied to the circumcircle of P EAF, we get LPEF = LPAF = LPAB. By Thales' Theorem applied to the circumcircle of PABC, we get LPAB = LPCB = LPCD. By Thales' Theorem applied to the circumcircle of P EC D, we get LPCD=LPED. Therefore, LP EF 1.4 = LP ED, which completes the proof. D More About Congruency The next theorem follows from ASA congruency together with the fact that the angle sum in a triangle is 180°. Theorem 1.4.1. (SAA Congruency) Two triangles are congruent if two angles and a side of one are congruent to two angles and the corresponding side of the other. In the figure below we have noncongruent triangles ABC and DEF. In these triangles, AB =DE, AC = DF, and LB = LE. This shows that, in general, SSA does not guarantee congruency. B ~ D C E 22 CONGRUENCY With further conditions we do get congruency: Theorem 1.4.2. (SSA + Congruency) SSA congruency is valid if the length of the side opposite the given angle is greater than or equal to the length of the other side. Proof. Suppose that in triangles ABC and DEF we have AB =DE, AC = DF, and LABC = LD E F and that the side opposite the given angle is the larger of the two sides. We will prove the theorem by contradiction. Assume that the theorem is false, that is, assume that BC ::f. EF; then we may assume that BC < EF. Let H be a point on EF so that EH = BC, as in the figure below. LJ D A B C E Then, 6ABC = 6DEH by SSS congruency. This means that DH = DF, so 6D H F is isosceles. Then LD FE = LD H F. Since we are given that DF ~ DE, the AngleSide Inequality tells us that LDEF~LDFE, and so it follows that LDEF Angle Inequality. ~ LDHF. However, this contradicts the External We must therefore conclude that the assumption that the theorem is false is incorrect, and so we can conclude that the theorem is true. D Since the hypotenuse of a right triangle is always the longest side, there is an immediate corollary: Corollary 1.4.3. (HSR Congruency) If the hypotenuse and one side of a right triangle are congruent to the hypotenuse and one side of another right triangle, the two triangles are congruent. MORE ABOUT CONGRUENCY 23 Counterexamples and Proof by Contradiction If we were to say that If ABCD is a rectangle, then AB = BC, you would most likely show us that the statement is false by drawing a rectangle that is not a square. When you do something like this, you are providing what is called a counterexample. In the assertion "If P, then Q," the statement P is called the hypothesis and the statement Q is called the conclusion. A counterexample to the assertion is any example in which the hypothesis is true and the conclusion is false. To prove that an assertion is not true, all you need to do is find a single counterexample. (You do not have to show that it is never true, you only have to show that it is not always true!) Exercise 1.4.4. For each of the following statements, provide a diagram that is a counterexample to the statement. 1. Given triangle ABC, if LABC = 60°, then ABC is isosceles. 2. Given that ABC is an isosceles triangle with AB = AC and with P and Q on side BC as shown in the picture, if BP = PQ = QC, then LBAP = LP AQ = LQAC. 3. In a quadrilateral ABCD, ABCD is a rectangle. if AB =CD and LBAD 6 B P Q C = LADC = 90°, then A proof by contradiction to verify an assertion of the form "If P, then Q" consists of the following two steps: (a) Assume that the assertion is false. This amounts to assuming that there is a counterexample to the assertion; that is, we assume that it is possible for the hypothesis P to be true while the conclusion Q is false. In other words, assume that it is possible for the hypothesis and the negative of the conclusion to simultaneously be true. (b) Show that this leads to a contradiction of a fact that is known to be true. In such circumstances, somewhere along the way an error must have been 24 CONGRUENCY made. Presuming that the reasoning is correct, the only possibility is that the assumption that the assertion is false must be in error. Thus, we must conclude that the assertion is true. 1.5 Perpendiculars and Angle Bisectors Two lines that intersect each other at right angles are said to be perpendicular to each other. The right bisector or perpendicular bisector of a line segment AB is a line perpendicular to AB that passes through the midpoint M of AB. I A+B perpendicular bisector The following theorem is the characterization of the perpendicular bisector. Theorem 1.5.1. (Characterization of the Perpendicular Bisector) Given different points A and B, the perpendicular bisector of AB consists of all points P that are equidistant from A and B. Proof. Let P be a point on the right bisector. Then in triangles PM A and PM B we have p PM=PM, LPMA =goo= LPMB, MA=MB, so triangles PM A and PM B are congruent by SAS. It follows that P A= P B. Conversely, suppose that P is some point such that P A = P B. Then triangles PM A and PM B are congruent by SSS. It follows that LP M A = LP M B, and since the sum of the two angles is 180°, we have LP M A = goo. That is, P is on the right bisector of AB. D Exercise 1.5.2. If m is the perpendicular bisector of AB, then A and B are on opposite sides of m. Show that if P is on the same side of m as B, then P is closer to B than to A. The following exercise follows easily from Pythagoras' Theorem. Try to do it without using Pythagoras' Theorem. Exercise 1.5.3. Show that the hypotenuse of a right triangle is its longest side. PERPENDICULARS AND ANGLE BISECTORS 25 Exercise 1.5.4. Let l be a line and let P be a point not on l. Let Q be the foot of the perpendicular from P to l. Show that Q is the point on l that is closest to P. Exercise 1.5.5. Let l be a line and let P be a point not on l. Show that there is at most one line through P perpendicular to l. Given a nonreflex angle LABC, a ray BD such that LABD angle bisector of LABC. = LCBD is called an A Given a line land a point P not on l, the distance from P to l, denoted d(P, l), is the length of the segment PQ where Q is the foot of the perpendicular from P to l. p d(P,l) l Q Theorem 1.5.6. (Characterization of the Angle Bisector) The angle bisector ofa nonrefiex angle consists of all points interior to the angle that are equidistant from the arms of the angle. 26 CONGRUENCY Proof. Let P be a point on the angle bisector. Let Q and R be the feet of the perpendiculars from P to AB and CB, respectively. Triangles PQB and PRB have the side P B in common, LPQB = LPRB and LPBQ = LPBR. Thus, the triangles are congruent by SAA, hence PQ P R. Therefore, P is equidistant from AB and C B. = Conversely, let P be a point that is equidistant from B A and BC. Let Q and R be the feet of the perpendiculars from P to AB and C B, respectively, so that PQ = P R. Thus, triangles PQ B and P RB are congruent by HSR, and it follows that LPBA = LPBQ = LPBR = A A LPBC. Hence, Pis on the angle bisector of LABC. D Inequalities in Proofs Before turning to construction problems, we list the inequalities that we have used in proofs and add one more to the list. 1. Triangle Inequality 2. Exterior Angle Inequality 3. AngleSide Inequality 4. Open Jaw Inequality This last inequality is given in the following theorem. PERPENDICULARS AND ANGLE BISECTORS 27 Theorem 1.5.7. (Open Jaw Inequality) Given two triangles 6.ABC and 6.DEF with AB = DE and BC L.ABC < L.D E F if and only if AC < D F, as in the figure. = EF. Then Proof. Suppose that x < y. Then we can build x in 6.DEF so that EG = AB. G can be inside or on the triangle. Here, we assume that G is outside 6.D E F, as in the figure below. F Now note that 6.ABC = 6.GEF by the SAS congruency theorem, and 6.EDG is isosceles with angles as shown. Also, s < L.DGF, since GE is interior to L.G, and s > L.GDF, since DF is interior to L.D. Therefore, L.DGF > s > L.GDF, and by the AngleSide Inequality, this implies that DF>GF=AC. Now suppose that AC < DF. Then exactly one of the following is true: x=y or x>y or x<y (this is called the law of trichotomy for the real number system). If x = y, then 6.ABC = 6.DEF by the SAS congruency theorem, which is a contradiction since we are assuming that AC < D F. If x > y, then by the first part of the theorem we would have AC also a contradiction. > DF, which is Therefore, the only possibility left is that x < y, and we are done. D 28 1.6 CONGRUENCY Construction Problems Although there are many ways to physically draw a straight line, the image that first comes to mind is a pencil sliding along a ruler. Likewise, the draftsman's compass comes to mind when one thinks of drawing a circle. To most people, the words straightedge and compass are synonymous with these physical instruments. In geometry, the same words are also used to describe idealized instruments. Unlike their physical counterparts, the geometric straightedge enables us to draw a line of arbitrary length, and the geometric compass allows us to draw arcs and circles of any radius we please. When doing geometry, you should regard the physical straightedge and compass as instruments that mimic the "true" or "idealized" instruments. There is a reason for dealing with idealized instruments rather than physical ones. Mathematics is motivated by a desire to look at the basic essence of a problem, and to achieve this we have to jettison any unnecessary baggage. For example, we do not want to worry about the problem of the thickness of the pencil line, for this is a drafting problem rather than a geometry problem. However, as we strip away the unnecessary limitations of the draftsman's straightedge and compass, the effect is to create versions of the instruments that behave somewhat differently from their physical counterparts. The idealized instruments are not "real," nor are the lines and circles that they draw. As a consequence, we cannot appeal to the properties of the physical instruments as verification for whatever we do in geometry. In order to work with idealized instruments, it is important to describe very clearly what they can do. The rules for the abstract instruments closely resemble the properties of the physical ones: Straightedge Operations A straightedge can be used to draw a straight line that passes through two given points. Compass Operations A compass can be used to draw an arc or circle centered at a given point with a given distance as radius. (The given distance is defined by two points.) These two statements completely describe how the straightedge and compass operate, and there are no further restrictions, nor any additional properties. For example, the most common physical counterpart of the straightedge is a ruler, and it is a fairly easy matter to place a ruler so that the line to be drawn appears to be tangent to a given physical circle. With the true straightedge, this operation is forbidden. If you wish to draw a tangent line, you must first find two points on the line and then use the straightedge to draw the line through these two points. CONSTRUCTION PROBLEMS 29 A ruler has another property that the straightedge does not. It has a scale that can be used for measuring. A straightedge has no marks on it at all and so cannot be used as a measuring device. We cannot justify our results by appealing to the physical properties of the instruments. Nevertheless, experimenting with the physical instruments sometimes leads to an understanding of the problem at hand, and if we restrict the physical instruments so that we only use the two operations described above, we are seldom led astray. Useful Facts in Justifying Constructions Recall that a rhombus is a parallelogram whose sides are all congruent, as in the figure on the right. A kite is a convex quadrilateral with two pairs of adjacent sides congruent. Note that the diagonals intersect in the interior of the kite. ·························>··· A dart is a nonconvex quadrilateral with two pairs of adjacent sides congruent. Note that the diagonals intersect in the exterior of the dart. Theorem 1.6.1. ( 1) The diagonals of a parallelogram bisect each other. (2) The diagonals of a rhombus bisect each other at right angles. (3) The diagonals (possibly extended) of a kite or a dart intersect at right angles. Basic Constructions The first three basic constructions are left as exercises. Exercise 1.6.2. To construct a triangle given two sides and the included angle. Exercise 1.6.3. To construct a triangle given two angles and the included side. Exercise 1.6.4. To construct a triangle given three sides. 30 CONGRUENCY Example 1.6.5. To copy an angle. Solution. Given LA and a point D, we wish to construct a congruent angle F DE. m A B D E Draw a line m through the point D. With center A, draw an arc cutting the arms of the given angle at Band C. With center D, draw an arc of the same radius cutting mat E. With center E and radius BC, draw an arc cutting the previous arc at F. Then, LF DE ::::: LA. Since triangles BAG and EDF are congruent by SSS, LBAC =LEDF. 0 Example 1.6.6. To construct the right bisector of a segment. Solution. Given points A and B, with centers A and B, draw two arcs of the same radius meeting at C and D. Then CD is the right bisector of AB. To see this, let M be the point where CD meets AB. First, we note that 6AC D ::::: 6BC D by SSS, so LAC D = LBC D. Then in triangles AC M and BC M we have AC=BC, LACM=LBCM, C M is common, so 6ACM::::: 6BCM by SAS. Then AM= BM and LAMC D = LBMC = 90°, which means that C M is the right bisector of AB. 0 CONSTRUCTION PROBLEMS 31 Example 1.6.7. To construct a perpendicular to a line from a point not on the line. p Solution. Let the point be P and the line be m. With center P, draw an arc cutting mat A and B. With centers A and B, draw two arcs of the same radius meeting at Q, where Q =/= P. Then PQ is perpendicular to m. Since by construction both P and Q are equidistant from A and B, both are on the right bisector of the segment AB, and hence PQ is perpendicular to m. D Example 1.6.8. To construct the angle bisector of a given angle. Solution. Let P be the vertex of the given angle. With center P, draw an arc cutting the arms of the angle at SandT. With centers SandT, draw arcs of the same radius meeting at Q. Then PQ is the bisector of the given angle. Since triangles S PQ and T PQ are congruent by SSS, LSPQ = LTPQ. P~r~ D Exercise 1.6.9. To construct a perpendicular to a line from a point on the line. 1.6.1 The Method of Loci The locus of a point that "moves" according to some condition is the traditional language used to describe the set of points that satisfy a given condition. For example, the locus of a point that is equidistant from two points A and B is the set of all points that are equidistant from A and B  in other words, the right bisector ofAB. The most basic method used to solve geometric construction problems is to locate important points by using the intersection of loci, which is usually referred to as the method of loci. We illustrate with the following: Example 1.6.10. Given two intersecting lines land m and a fixed radius r, construct a circle of radius r that is tangent to the two given lines. 32 CONGRUENCY Solution. It is often useful to sketch the expected solution. We refer to this sketch as an analysis figure. The more accurate the sketch, the more useful the figure. In the analysis figure you should attempt to include all possible solutions. The analysis figure for Example 1.6.1 0 is as follows, where l and m are the given lines intersecting atP. m The analysis figure indicates that there are four solutions. The constructions of all four solutions are basically the same, so in this case it suffices to show how to construct one of the four circles. Since we are given the radius of the circle, it is enough to construct 0, the center of circle C. Since we only have a straightedge and a compass, there are only three ways to construct a point, namely, as the intersection of • two lines, • two circles, or • a line and a circle. The center 0 of circle C is equidistant from both l and m and therefore lies on the following constructible loci: 1. an angle bisector, 2. a line parallel to l at distance r from l, 3. a line parallel tom at distance r from m. m Any two of these loci determine the point 0. SOLUTIONS TO SELECTED EXERCISES 33 Having done the analysis, now write up the solution: 1. Construct linen parallel to l at distance r from l. 2. Construct line k parallel to m at distance r from m. 3. Let 0 = n n k. 4. With center 0 and radius r, draw the circle C(0, r). k m D 1.7 Solutions to Selected Exercises Solution to Exercise 1.2.8 1. Statement: Given triangle ABC, if L.ABC is a right angle, then AB 2 + BC2 = AC2 . Converse: Given triangle ABC, if AB 2 + BC 2 = AC2 , then L.ABC is a right angle. Both the statement and its converse are true. 34 CONGRUENCY = CD and AD= BC. Converse: If AB = CD and AD= BC, then ABCD is a parallelogram. 2. Statement: If ABCD is a parallelogram, then AB The statement is true and the converse is false. 3. Statement: If ABC D is a convex quadrilateral, then ABC D is a rectangle. Converse: If ABCD is a rectangle, then ABC D is a convex quadrilateral. The statement is false and the converse is true. 4. Statement: Given quadrilateral ABCD, if AC =1 BD, then ABCD is not a rectangle. Converse: Given quadrilateral ABCD, if ABCD is not a rectangle, then AC =/: BD. The statement is true and its converse is false. c Solution to Exercise 1.3.9 In the figure below, we have a = 30 (isosceles triangle). Thus, by Thales' Theorem, (3 = 2(a + 40) = 140, so that () = 180  (3 = 40. 35 SOLUTIONS TO SELECTED EXERCISES Solution to Exercise 1.3.14 Suppose that the quadrilateral ABC D has an inscribed circle that is tangent to the sides at points P, Q, R, and S, as shown in the figure. D c Since the tangents to the circle from an external point have the same length, then and AP = AQ, P B = BS SC=CR, RD=QD, so that AB+CD = AP+PB+CR+RD = AQ+BS+SC+QD (AQ + QD) =AD+BC. = + (BS + SC) Conversely, suppose that AB + CD = AD + BC, and suppose that the extended sides AD and BC meet at X. Introduce the incircle of L:.DXC, that is, the circle internally tangent to each of the sides of the triangle, and suppose that it is not tangent to AB. Let E and F be on sides AD and BC, respectively, such that EF is parallel to AB and tangent to the incircle, as in the figure. D "' ~ X B F c 36 CONGRUENCY Note that since 6X EF"' 6X AB with proportionality constant k > 1, EF > AB, and since the quadrilateral DEFC has an inscribed circle, then by the first part of the proof we must have AB+CD < EF+CD = DE+CF < AD+BC, which is a contradiction. When the side AB intersects the circle twice, a similar argument also leads to a contradiction. Therefore, if the condition AB+CD = AD+BC holds, then the incircle of 6DXC must also be tangent to AB, and ABCD has an inscribed circle. The case when ABCDis a parallelogram follows in the same way. First, we construct a circle that is tangent to three sides of the parallelogram. The center of the circle must lie on the line parallel to the sides AD and BC and midway between them. Let 2r be the perpendicular distance between AD and BC. Then the center must also lie on the line parallel to the side CD and at a perpendicular distance r from CD, as in the figure. B F As before, suppose that the circle is not tangent to AB, and let E and F be on sides AD and BC, respectively, such that EF is parallel to AB and tangent to the circle, as in the figure above. Now, AB = EF, and since the quadrilateral DEFC has an inscribed circle, by the first part of the proof we must have AB+CD = EF+CD = DE+CF < AD+BC, which is a contradiction. When the side AB intersects the circle twice, a similar argument also leads to a contradiction. Therefore, if the condition AB+CD = AD+BC holds, then the circle must also be tangent to AB, and the parallelogram ABC D has an inscribed circle. SOLUTIONS TO SELECTED EXERCISES 37 Solution to Exercise 1.4.4 1. A 2. The assertion is always false. Solution to Exercise 1.5.2 In the figure below, we have P B < PQ + QB = PQ + QA = P A. p Solution to Exercise 1.5.3 Here are two different solutions. 1. The right angle is the largest angle in the triangle (otherwise the sum of the three angles of the triangle would be larger than 180°). Since the hypotenuse is opposite the largest angle, it must be the longest side. 2. In the figure below, suppose that B is the right angle. c 38 CONGRUENCY Extend AB beyond B to D so that BD = AB. Then Cis on the right bisector of AD, so 1 1 AC = 2(AC +CD) > 2AD = AB. This shows that AC > AB, and in a similar fashion it can be shown that AC > CB. Solution to Exercise 1.5.4 Let Q be the foot of the perpendicular from P to the line l, and let R be any other point on l. Then !:::.PQR is a right triangle, and by Exercise 1.5.3, P R > PQ. p l R 1.8 Problems 1. Prove that the internal and external bisectors of the angles of a triangle are perpendicular. 2. Let P be a point inside C(0, r) with P # 0. Let Q be the point where the ray ; OP meets the circle. Use the Triangle Inequality to show that Q is the point on the circle that is closest to P. 3. Let P be a point inside !:::.ABC. Use the Triangle Inequality to prove that AB+BC > AP+PC. 4. Each of the following statements is true. State the converse of each statement, and if it is false, provide a figure as a counterexample. (a) If !:::.ABC= !:::.DEF, then LA= LD and LB = LE. =goo. If ABCD is a rectangle, then LA = LB = LC = goo. (b) If ABCD is a rectangle, then LA= LC (c) 5. Given the isosceles triangle ABC with AB = AC, let D be the foot of the perpendicular from A to BC. Prove that AD bisects LBAC. PROBLEMS 39 6. Show that if the perpendicular from A to BC bisects L_BAC, then LABC is isosceles. 7. D is a point on BC such that AD is the bisector of L_A. Show that L_ADC 90+ L_B L_C  2 . 8. Construct an isosceles triangle ABC, given the unequal angle L_A and the length of the side BC. 9. Construct a right triangle given the hypotenuse and one side. 10. Calculate() in the following figure. (not drawn to scale) 11. Let Q be the foot of the perpendicular from a point P to a line l. Show that Q is the point on l that is closest toP. 12. Let P be a point inside C(0, r) with P 1 0. Let Q be the point where the ray ) PO meets the circle. Show that Q is the point of the circle that is farthest from P. 13. Let ABCD be a simple quadrilateral. Show that ABCD is cyclic if and only if the opposite angles sum to 180°. 14. Draw the locus of a point whose sum of distances from two fixed perpendicular lines is constant. 40 CONGRUENCY 15. Given a circle C(P, s), a line l disjoint from C(P, s), and a radius r (r > s), construct a circle of radius r tangent to both C(P, s) and l. Note: The analysis figure indicates that there are four solutions. CHAPTER2 CONCURRENCY 2.1 Perpendicular Bisectors This chapter is concerned with concurrent lines associated with a triangle. A family of lines is concurrent at a point P if all members of the family pass through P. In preparation, we need a few additional facts about parallel lines. Theorem 2.1.1. Let h and l 2 be parallel lines, and suppose that lines with m1 ..l h and m2 ..l h. Then m1 and m2 are parallel. m 1 and Classical Geometry: Euclidean, Transformational, Inversive and Projective, First Edition. By I. E. Leonard, J. E. Lewis, A. C. F. Liu, G. W. Tokarsky Copyright© 2014 John Wiley & Sons, Inc. m2 are 41 42 CONCURRENCY Q s p w Proof. Let P be the point h n m 1 and let Q be the point l 2 n m 2 . By the parallel postulate, h is the only line through P parallel to l 2 , and so m 1 is not parallel to b and consequently must meet l2 at some pointS. In other words, m 1 is a transversal for the parallel lines h and l 2 . Since the sum of the adjacent interior angles is 180°, it follows that b must be perpendicular to m 1 . But l2 is also perpendicular to m2 and so m 1 and m2 must be parallel. D In the above proof, if we interchange h and m 1 and l 2 and m2, we have: Corollary 2.1.2. Suppose that h _l m1 and l2 and only ifm1 and m2 are parallel. _l m2. Then h and l2 are parallel if One of the consequences of this corollary is that if two line segments intersect, then their perpendicular bisectors must also intersect. This fact is crucial in the following theorem, the proof of which also uses the fact that the right bisector of a segment can be characterized as being the set of all points that are equidistant from the endpoints of the segment. Theorem 2.1.3. The perpendicular bisectors ofthe sides ofa triangle are concurrent. A B c ANGLE BISECTORS 43 Proof. According to the comments preceding the theorem, the perpendicular bisectors of AB and AC meet at some point 0, as shown in the figure. It is enough to show that 0 lies on the perpendicular bisector of BC. We have = OA (since 0 OA = OC (since 0 OB Therefore OB is on the perpendicular bisector of AB), is on the perpendicular bisector of AC). = OC, which means that 0 is on the perpendicular bisector of BC. D The proof shows that the point 0 is equidistant from the three vertices, so with center 0 we can circumscribe a circle about the triangle. The circle is called the circumcircle, and the point 0 is called the circumcenter of the triangle. Exercise 2.1.4. In the figure above, the circumcenter is interior to the triangle. In what sort of a triangle is the circumcenter I. on one of the edges of the triangle? 2. outside the triangle? 2.2 Angle Bisectors Theorem 2.2.1. For a circle C(0, r): I. The right bisector of a chord ofC(O, r) passes through 0. 2. If a given chord is not a diameter, the line joining 0 to the midpoint of the chord is the right bisector of the chord. 3. The line from 0 that is perpendicular to a given chord is the right bisector of the chord. 4. A line is tangent to C(0, r) at a point P radiusOP. I. 2. if and only if it is perpendicular to the 3. 4. 44 CONCURRENCY Theorem 2.2.2. The internal bisectors of the angles of a triangle are concurrent. A c B Proof. In 6ABC, let the internal bisectors of LB and LC meet at I, as shown above. We will show that I lies on the internal bisector of LA. We have d(I, AB) = d(I, BC), since I lies on the internal bisector of LB, and d(I, BC) = d(I, AC), since I lies on the internal bisector of LC. Hence, d(I, AB) = d(I, AC), which means that I is on the internal bisector of LA. 0 The proof of Theorem 2.2.2 shows that if we drop the perpendiculars from I to the three sides of the triangle, the lengths of those perpendiculars will all be the same, say r. Then the circle C(I, r) will be tangent to all three sides by Theorem 2.2.1. That is, I is the center of a circle inscribed in the triangle. This inscribed circle is called the incircle of the triangle, and I is called the incenter of the triangle. A We will next prove a theorem about the external angle bisectors. First, we need to show that two external angle bisectors can never be parallel. ANGLE BISECTORS 45 Proposition 2.2.3. Each pair of external angle bisectors of a triangle intersect. Proof. Referring to the diagram, 28 =a + ')', 2¢ = f3 + ')'. Adding and rearranging: 2(8 + ¢) = 180 + 'Y · Since 'Y < 180, it follows that()+ 4> < 180, so the external angle bisectors cannot be parallel. D Theorem 2.2.4. The external bisectors of two of the angles of a triangle and the internal bisector of the third angle are concurrent. B C Proof. Let the two external bisectors meet atE, as shown in the figure . It is enough to show that E lies on the internal bisector of L_B . Now, d(E, AB) = d(E , AC), since E lies on the external bisector of LA, and d(E , AC) = d(E, BC) , since E lies on the external bisector of LC. Hence, d(E, AB) means that E is on the internal bisector of L_B . = d(E, BC) , which D The point of concurrency E is called an excenter of the triangle. Since E is equidistant from all three sides (some extended) of the triangle, we can draw an excircle tangent to these sides. Every triangle has three excenters and three excircles. 46 2.3 CONCURRENCY Altitudes A line passing through a vertex of a triangle perpendicular to the opposite side is called an altitude of the triangle. To prove that the altitudes of a triangle are concurrent, we need some facts about parallelograms. A parallelogram is a quadrilateral whose opposite sides are parallel. A parallelogram whose sides are equal in length is called a rhombus. Squares and rectangles are special types of parallelograms. Theorem 2.3.1. In a parallelogram: (I) Opposite sides are congruent. (2) Opposite angles are congruent. (3) The diagonals bisect each other. 0 CJ p Proof. We will prove (1) and (2). Given parallelogram ABCD, as in the figure below, we will show that 6ABC = 6CDA. Diagonal AC is a transversal for parallel lines AB and CD, and so L.BAC and L.DCA are opposite interior angles for the parallel lines. So we have L.BAC = L.DCA. Treating AC as a transversal for AD and CB, we have L.BCA = L.DAC. ALTITUDES 47 Since AC is common to 6.BAC and !::,DCA, ASA implies that they are congruent. Consequently, AB =CD and BC = DA, proving (1). Also, LBAD =a+ {3 = LBCD and LADC and ABC are congruent, proving (2). = LABC because triangles ADC D Exercise 2.3.2. Prove statement (3) of Theorem 2.3.1. Theorem 2.3.3. If the diagonals ofa quadrilateral bisect each other, then the quadrilateral is a parallelogram. Exercise 2.3.4. Prove Theorem 2.3.3. Theorem 2.3.3 tells us that statement (3) of Theorem 2.3.1 will guarantee that the quadrilateral is a parallelogram. However, neither statement (1) nor statement (2) of Theorem 2.3.1 is by itself enough to guarantee that a quadrilateral is a parallelogram. For example, a nonsimple quadrilateral whose opposite sides are congruent is not a parallelogram. However, if the quadrilateral is a simple polygon, then either statement (1) or (2) is sufficient. As well, there is another useful condition that can help determine if a simple quadrilateral is a parallelogram: Theorem 2.3.5. A simple quadrilateral is a parallelogram statements are true: if any of the following ( 1) Opposite sides are congruent. (2) Opposite angles are congruent. (3) One pair of opposite sides is congruent and parallel. Proof. We will justify case (1), leaving the others to the reader. Suppose that ABC D is the quadrilateral, as in the figure on the right. Since ABC D is simple, we may suppose that the diagonal AC is interior to the quadrilateral. Since the opposite sides of the quadrilateral are congruent, the SSS congruency condition implies that !::,ABC= !::,CDA. This in tum implies that the alternate interior angles LB AC and LDC A are congruent and also that the alternate interior angles LBC A and LD AC are congruent. The fact that the edges are parallel now follows from the wellknown facts about parallel lines. D 48 CONCURRENCY The next theorem uses a clever trick. It embeds the given triangle in a larger one in such a way that the altitudes of the given triangle are right bisectors of the sides of the larger one. Theorem 2.3.6. The altitudes of a triangle are concurrent. Proof. Given !:::,.ABC, we embed it in a larger triangle !:::,.DE F by drawing lines through the vertices of !:::,.ABC that are parallel to the oppositesidessothatABCE, ACBF, andCABD are parallelograms, as in the diagram on the right. Clearly, A, B, and C are midpoints of the sides of !:::,.DEF, and an altitude of !:::,.ABC is a perpendicular bisector of a side of !:::,.DE F. However, since the perpendicular bisectors of !:::,.DEF are concurrent, so are the altitudes of !:::,.ABC. 0 The point of concurrency of the altitudes is called the orthocenter of the triangle, and it is usually denoted by the letter H. In the proof, the orthocenter of !:::,.ABC is the circumcenter of !:::,.DEF. Note that the orthocenter can lie outside the triangle (for obtuseangled triangles) or on the triangle (for rightangled triangles). The same proof also works for these types of triangles. Exercise 2.3.7. The figure above shows a triangle whose orthocenter is interior to the triangle. Give examples of triangles where: 1. The orthocenter is on a side of the triangle. 2. The orthocenter is exterior to the triangle. 2.4 Medians A median of a triangle is a line passing through a vertex and the midpoint of the opposite side. Exercise 2.4.1. Show that in an equilateral triangle ABC the following are all the same: 1. The perpendicular bisector of BC. 2. The bisector of LA. 3. The altitude from vertex A. 4. The median passing through vertex A. MEDIANS 49 The next theorem, which is useful on many occasions, is also proved by using the properties of a parallelogram. Theorem 2.4.2. (The Midline Theorem) If P and Q are the respective midpoints of sides AB and AC of triangle ABC, then PQ is parallel to BC and PQ = BC /2. A Proof. First, extend PQ toT so that PQ = QT, as in the figure. Then ATCP is a parallelogram, because the diagonals bisect each other. This means that TC is parallel to and congruent to AP. Since Pis the midpoint of AB, it follows that TC is parallel to and congruent to B P, and so T C B P is also a parallelogram. From this we can conclude that PT is parallel to and congruent to BC; that is, PQ is parallel to BC and half the length of BC. D Lemma 2.4.3. Any two medians of a triangle trisect each other at their point of intersection. A B Proof. Let the medians BE and CF intersect at G, as shown in the figure. Draw FE. By the Midline Theorem, FE is parallel to BC and half its length. Let P be the midpoint of GB and let Q be the midpoint of GC. ·Then, again by the Midline Theorem, PQ is parallel to BC and half its length. Since FE and PQ are parallel and equal in length, EF PQ is a parallelogram, and so the diagonals of E F PQ bisect each other. It follows that FG = GQ = QC and EG = G P = P B, which proves the lemma. D 50 CONCURRENCY There are two different points that trisect a given line segment. The point of intersection of the two medians is the trisection point of each that is farthest from the vertex. Theorem 2.4.4. The medians of a triangle are concurrent. Proof. Let the medians be AD, BE, and CF. Then BE and CF meet at a point G for which EG/EB = 1/3. Also, AD and BE meet at a point G' for which EG' / EB = 1/3. Since both G and G' are between Band E, we must have G = G', which means that the three medians are concurrent. D The point of concurrency of the three medians is called the centroid. The centroid always lies inside the triangle. A thin triangular plate can be balanced at its centroid on the point of a needle, so physically the centroid corresponds to the center of gravity. The partial converses of the Midline Theorem are useful: Theorem 2.4.5. Let P be the midpoint of side AB of triangle ABC, and let Q be a point on AC such that PQ is parallel to BC. Then Q is the midpoint of AC. Proof. Let Q' be the midpoint of AC, then PQ' II BC. Since there is only one line through P parallel to BC, the lines PQ and PQ' must be the same, so the points Q and Q' are also the same. D 2.5 Construction Problems The two basic construction problems associated with parallel lines are constructing a parallelogram given two of its adjacent edges (that is, completing a parallelogram) and constructing a line through a given point parallel to a given line. Once we have solved the first problem, the second one is straightforward. Example 2.5.1. To construct a parallelogram given two adjacent edges. Solution. Given edges AB and AC, with center B and radius AC, draw an arc. With center C and radius AB, draw a second arc cutting the first at D on the same side of AC as B. Then ABDC is the desired parallelogram. c__  ~ '' ' ' ' A L         1 'B D CONSTRUCTION PROBLEMS 51 Example 2.5.2. To construct a line parallel to a given line through a point not on the line. Solution. With center P, draw an arc cutting the given line at Q and S. With the same radius and center S, draw a second arc. With center P and radius QS, draw a third arc cutting the second at T. Then PT is the desired line, because PQST is a parallelogram. p ~T ·/' D Example 2.5.3. Given a circle with center 0 and a point P outside the circle, construct the lines through P tangent to the circle. Solution. Draw the line OP and construct the right bisector of OP, obtaining the midpoint M of OP. With center M and radius M P, draw a circle cutting the given circle at S and T. Then P S and PT are the desired tangent lines. Note that angles OSP and OTP are angles in a semicircle, so by Thales' Theorem both are right angles. Thus, PS and PT are perpendicular to radii OS and OT, respectively, and so must be tangent lines by Theorem 2.2.1. D Thales' Locus Given a segment AB and an angle (}, the set of all points P such that LAP B = (} forms the union oftwo arcs of a circle, which we shall call Thales' Locus, shown in the figure on the following page. 52 CONCURRENCY 8 acute (J obtuse 8=90 To help us construct Thales' Locus, we need the following useful theorem. Theorem 2.5.4. Let AB be a chord of the circle C(0, r) and let P be a point on the line tangent to the circle, as shown in the diagram below. Then L.AOB = 2L.P AB. Proof. Referring to the diagram, drop the perpendicular 0 M from 0 to AB. Then L.AOB = 2L.AOM = 2a. Now, () = 90 ¢ = a, and it follows that LAO B = 2L.P AB. D Note. In a similar way, it can be shown that the reflex angle AO B is twice the size of L.QAB. CONSTRUCTION PROBLEMS Example 2.5.5. Given a segment AB and an angle for the given data: e < 90°, construct Thales' Locus L Data: A 53 B Solution. Here is one method of doing this construction. 1. Copy angle (} to point A so that AB is one arm of the angle, and l is the line containing the other arm. 2. Construct the right bisector mofAB. L 3. Construct the line t through A perpendicular to l. Let 0 be the point t n m. 4. Draw C(O, OA). Then one of the arcs determined by C(O, OA) is partofThales' Locus for the given data. By Theorem 2.5.4, LAOB = 2(}, so LAX B at the circumference is of size e. D Example 2.5.6. Construct a triangle ABC given the size and C, and the length of the altitude h from A. e of LA, the vertices B Solution. We give a quick outline of the solution. 1. Construct Thales' Locus for BC and e. 2. Construct the perpendicular BX to BC so that BX =h. 3. Through X, draw the line l parallel to BC cutting the locus at a point A. Then ABC is the desired triangle. D 54 CONCURRENCY 2.6 Solutions to the Exercises Solution to Exercise 2.1.4 0 In an acuteangled triangle, the circumcenter is always interior to the triangle. 1. In a right triangle, the circumcenter is always on the hypotenuse. 2. In an obtuseangled triangle, the circumcenter is always outside the triangle. Solution to Exercise 2.3.2 A In triangles AB P and CD P we have a="(, (3 = 8, because AB II CD, and AB=CD, by statement (1) of Theorem 2.3.1. So, 6.ABP = 6.CDP by ASA, and it follows that AP = CP and BP = DP. SOLUTIONS TO THE EXERCISES 55 Solution to Exercise 2.3.4 A In triangles AB P and CD P we have AP= CP, L_APB = L_CPD since they are vertically opposite angles, and BP=DP. So triangles ABP and CDP are congruent, and thus L_PAB implies that AB II CD. Similarly, AD I L_PCD, which CB, which shows that ABCD is a parallelogram. Solution to Exercise 2.3.7 In an acuteangled triangle, the orthocenter is always interior to the triangle. 1. In a right triangle, the orthocenter is always the vertex of the right angle. 2. In an obtuseangled triangle, the orthocenter is always outside the triangle. 56 CONCURRENCY Solution to Exercise 2.4.1 A c B We will show that the line through the vertex A and the midpoint D of BC is simultaneously the perpendicular bisector of BC, the bisector of LA, the altitude from A, and the median from A. = AC, the point A is equidistant from Band C and so A is on the right bisector of BC. It follows that AD is the right bisector of BC. 1. Since AB 2. Triangles ADB and ADC are congruent by SSS, so LDAB is, AD is the bisector of LA. = LDAC; that 3. Since AD ..l BC, by statement 1 above, AD is an altitude of 6ABC. 4. Since Dis the midpoint of BC, AD is a median of 6ABC. 2.7 Problems 1. BE and CF are altitudes of 6ABC, and M is the midpoint of BC. Show that ME=: MF. 2. BE and CF are altitudes of L!.ABC, and EF is parallel to BC. Prove that 6ABC is isosceles. 3. The perpendicular bisector of side BC of 6ABC meets the circumcircle at D on the opposite side of BC from A. Prove that AD bisects LBAC. 4. Given L!.ABC with incenter I, prove that LBIC = 90 + ~LBAC. Note. This is an important property of the incenter that will prove useful later. PROBLEMS 57 5. In the given figure, calculate the sizes of the angles marked a and (3. c A 6. In 6ABC, L_BAC = 100° and L_ABC a median. Find L_CDE. = 50°. AD is an altitude and BE is 7. Segments P S and PT are tangent to the circle at S and T. Show that (a) PS = PT and (b) ST _lOP. T 8. Construct 6ABC given the side BC and the lengths hb and he of the altitudes from Band C, respectively. 9. Construct 6ABC given the side BC, the length hb of the altitude from B, and the length ma of the median from A. I 0. Construct 6ABC given the length of the altitude h from A and the length of the sides b and c. Here, b is the side opposite L_B, and c is the side opposite L_C. 11. Construct triangle ABC given BC, an angle (3 congruentto L_B, and the length t ofthe median from B. 58 CONCURRENCY 12. Given a point P inside angle ABC as shown below, construct a segment XY with endpoints in AB and CB such that Pis the midpoint of XY. A . p BL C 13. Given segments AB and CD, which meet at a point P off the page, construct the bisector of LP. All constructions must take place within the page. 14. Let AB be a diameter of a circle. Show that the points where the right bisector of AB meet the circle are the points of the circle that are farthest from AB. 15. M is the midpoint of the chord AB of a circle C(O, r). Show that if a different chord CD contains M, then AB < CD. (You may use Pythagoras' Theorem.) 16. ABC D is a nonsimple quadrilateral. P, Q, R, and S are the midpoints of AB, BC, CD, and DA, respectively. Show that PQRS is a parallelogram. 17. A regular polygon is one in which all sides are equal and all angles are equal. Show that the vertices of a regular convex polygon lie on a common circle. CHAPTER 3 SIMILARITY 3.1 Similar Triangles The word similar is used in geometry to describe two figures that have identical shapes but are not necessarily the same size. A working definition of similarity can be obtained in terms of angles and ratios of distances. Two polygons are similar if corresponding angles are congruent and the ratios of corresponding sides are equal. Classical Geometry: Euclidean, Transformational, Inversive and Projective, First Edition. By I. E. Leonard, J. E. Lewis, A. C. F. Liu, G. W. Tokarsky Copyright© 2014 John Wiley & Sons, Inc. 59 60 SIMILARITY Notation. We use the symbol rv to denote similarity. Thus, we write ABCDE rv QRSTU to denote that the polygons ABC DE and Q RSTU are similar. As with congruency, the order of the letters is important. To say that !::,.ABC rv 6.DEF means that LA=:LD, and AB DE LB = LE, BC LC=:LF, AC = EF = DF = k' where k is a positive real number. A D ~ E B F The constant k is called the proportionality constant or the magnification factor. If k > 1, triangle ABC is larger than triangle DEF; if 0 < k < 1, triangle ABC is smaller than triangle DEF; and if k = 1, the triangles are congruent. Note that congruent figures are necessarily similar, but similar figures do not need to be congruent. 3.2 Parallel Lines and Similarity There is a close relationship between parallel lines and similarity. Lemma 3.2.1. In 6.ABC, suppose that D and E are points of AB and AC, respectively, and that DE is parallel to BC. Then AD DB AE EC. Figures (i), (ii), and (iii) on the following page illustrate the three different possibilities that can occur. PARALLEL LINES AND SIMILARITY 61 A B D/~E (ii) (i) (iii) The proof of this lemma uses some simple facts about areas: Theorem 3.2.2. Given parallel lines land m and two triangles each with its base on one line and its remaining vertex on the other, the ratio of the areas of the triangles is the ratio of the lengths of their bases. A m B D C E F For the diagram above, the theorem says that [ABC] [DEF] BC EF' Note that we use square brackets to denote area; that is, [XY Z] is the area of 6XY Z. There are two special cases worth mentioning, and these are illustrated by the figures below: A B ~ C [ABC] [ACD] D BC CD A D B C [ABC] = [DBC] We sketch the proof of Lemma 3.2.1 for case (i). The proof of the other cases is very much the same. 62 SIMILARITY Proof. In case (i), the line DE enters the triangle ABC through side AB and so must exit either through vertex C or through one of the other two sides. 1 Since DE is parallel to BC, the line DE cannot pass through vertex Cor any other point on side BC. It follows that DE must exit the triangle through side AC. Insert segments BE and CD and use the previously cited facts about areas of triangles to get A [ADE] [BDE] AD BD [ADE] [CDE] and AE CE. Since [BDE] = [CDE], we must have AD DB B AE EC' ~ C D A useful extension of the lemma is the following: Theorem 3.2.3. Parallel projections preserve ratios. Suppose that l, m, and n are parallel lines that are met by transversals t and t' at points A, B, C and A', B', C', respectively. Then AB BC A'B' B'C'' Proof. Draw a line through A parallel to t' meeting m at B" and n at C". Then AA' B' B" and B" B' C' C" are parallelograms, so AB" = A' B' and B" C" = B' C'. Applying Lemma 3.2.1 to triangle ACC", we have AB BC and the theorem follows. 1This statement is known as Pasch's Axiom. AB" B"C"' D PARALLEL LINES AND SIMILARITY 63 Next is the basic similarity theorem for triangles: Theorem 3.2.4. In 6.ABC, suppose that DE is parallel to BC. If D and E are points of AB and AC, with D being neither A nor B nor C, then 6.ABC rv 6.ADE. Proof. As in the case of Lemma 3.2.1, there are three cases to consider: (1) Dis between A and B. ___. (2) Dis on the ray AB beyond B. ___. (3) Dis on the ray BA beyond A. We will prove the theorem for case (1). The proofs for the other cases are almost identical. Because of the properties of parallel lines, the corresponding angles of ABC and AD E are equal, so it remains to show that the ratios of the corresponding sides are equal. B A F A C By Lemma 3.2.1, we have AD DB AE EC' DB AD EC AE' which implies that and this in tum implies that so that AD+DB AD AE+EC AE and therefore AB AD AC AE' Now, through E, draw EF parallel to AB, with F on BC. Using an argument similar to the one above, we get AC AE BC BF' 64 SIMILARITY Since DE= BF (because BDEF is a parallelogram), we get AC AE BC DE Thus, AB AC BC AD AE DE' which completes the proof that triangles ABC and ADE are similar. D There are other conditions that allow us to conclude that two triangles are similar, but there are many occasions where Theorem 3.2.4 is the most appropriate one to use. Congruency versus Similarity Congruency and similarity are both equivalence relations  both relations are reflexive, symmetric, and transitive. Both are reflexive: !:::.ABC:=!:::.ABC. !:::.ABC "' !:::.ABC. Both are symmetric: If !:::.ABC= !:::.DEF, then !:::.DEF !:::.ABC. If !:::.ABC"' !:::.DEF, then !:::.DEF"' !:::.ABC. = Both are transitive: If !:::.ABC= !:::.DEF and !:::.DEF = !:::.GHI, then !:::.ABC= !:::.GHI. If !:::.ABC"' !:::.DEF and !:::.DEF"' !:::.GHI, then !:::.ABC"' !:::.GHI. 3.3 Other Conditions Implying Similarity According to the definition, two triangles are similar if and only if the three angles are congruent and the three ratios of the corresponding sides are equal. As with congruent triangles, it is not necessary to check all six items. Here are some of the conditions that will allow us to verify that triangles are similar without checking all SIX. 65 OTHER CONDITIONS IMPLYING SIMILARITY Theorem 3.3.1. (AAA or AngleAngleAngle Similarity) Two triangles are similar if and only if all three corresponding angles are congruent. Exercise 3.3.2. Show that two quadrilaterals need not be similar even corresponding angles are congruent. if all their Theorem 3.3.3. (sAs or sideAngleside Similarity) /fin LABC and LDEF we have AB DE AC DF and LA= LD, then LABC and LDEF are similar. Theorem 3.3.4. (sss or sidesideside Similarity) Triangles ABC and DEF are similar if and only if AB DE BC EF AC DF' The lowercase letter s in sAs and sss is to remind us that the sides need only be in proportion rather than congruent, while the uppercase letter A is to remind us that the angles must be congruent. It is worth mentioning that since there are 180° in a triangle, the AAA similarity condition is equivalent to: Theorem 3.3.5. (AA Similarity) Two triangles are similar congruent. if and only if two of the three corresponding angles are Proofofthe AAA similarity condition. Make a congruent copy of one triangle so that it shares an angle with the other, that is, so that two sides of the one triangle fall upon two sides of the other. The congruency of the angles then guarantees that the third sides are parallel, and the proof is completed by applying Theorem 3.2.4. D LJ A B C 66 SIMILARITY The proof for the sAs similarity condition uses the converse of Lemma 3 .2.1. Lemma 3.3.6. Let P and Q be points on AB and AC with P between A and B and Q between A and C. If AP PB AQ QC' then PQ is parallel to BC. Proof. Through P, draw a line P R parallel to BC with Ron AC. Then, since P R enters triangle ABC through side AB, it must exit the triangle through side AC or side BC. Since P R is parallel to BC, it follows that R is between A and C. B By Lemma 3.2.1, A AP PB A C AR RC' so it follows that However, given the positive number k, there is only one point X between A and C such that AX XC =k, and so it follows that R = Q, and so PQ = P R, showing that PQ is parallel to BC. D Proof of the sAs similarity condition. Suppose that in triangles ABC and DEF we have LA= LD and that AB AC DE= DF =k. We may assume that k > 1, that is, that AB > DE and AC > DF. Cut off a segment AP on AB so that AP = DE. Cut off a segment AQ on AC so that AQ=DF. EXAMPLES 67 Then it follows that AB AP A AC AQ' and subtracting 1 from both sides of the equation gives us PB AP QC AQ. It now follows from Lemma 3.3.6 that PQ is parallel to BC, and Theorem 3.2.4 implies that l:J.ABC"' l:J.APQ. Since l:J.APQ = l:J.DEF (by the SAS congruency condition), it follows that l:J.ABC"' l:J.DEF. 0 Exercise 3.3.7. Prove that two triangles are similar if they satisfy the sss similarity condition. 3.4 Examples Pythagoras' Theorem Theorem 3.4.1. (Pythagoras' Theorem) If two sides of a right triangle have lengths a and band the hypotenuse has length c, then a 2 + b2 = c2 . Proof. In the figure below, drop the perpendicular CD to the hypotenuse, and let AD = p and BD = q. c A ~ D c B From the AA similarity condition, l:J.CBD"' l:J.ABC and l:J.AC D "' l:J.ABC, 68 SIMILARITY implying that a q c a b and c b' p or, equivalently, that and from which it follows that a2 and since q + p + b2 = cq + cp = c(q + p), = c we have a 2 + b2 = c2 . D Theorem 3.4.2. (Converse of Pythagoras' Theorem) Let ABC be a triangle. If then LC is a right angle. Exercise 3.4.3. Prove the converse of Pythagoras' Theorem. Apollonius' Theorem As an application of Pythagoras' Theorem, we prove the following result. Theorem 3.4.4. (Apollonius' Theorem) Let M be the midpoint of the side BC of triangle ABC. Then Proof. Let AD be the altitude on the base BC, and assume that D lies between M and C. A B M D C EXAMPLES 69 By Pythagoras' Theorem, AB 2 + AC2 BD 2 + 2AD 2 + CD 2 = = (BM + MD) 2 + (BM MD) 2 + 2AD 2 = 2BM 2 +2M D 2 + 2AD 2 = 2BM 2 + 2AM 2 . For other positions of D, the argument is essentially the same. D Example 3.4.5. Use Apollonius' Theorem to prove that in .6.ABC, ifma =AM is the median from the vertex A, then where a = BC, b = AC, and c = AB, as in the figure. A b c B a M Solution. From Apollonius' Theorem, we have that is, so that Therefore, D Stewart's Theorem A related theorem is the following, interesting in its own right, which allows us to express the lengths of the internal angle bisectors of a triangle in terms of the lengths of the sides. 70 SIMILARITY Theorem 3.4.6. (Stewart's Theorem) In 6.ABC, if Dis any point internal to the segment BC, d = AD, m = BD, n =DC, b = AC, c = AB, and a= BC, as in the figure, then c2 · n + b2 · m = a · (d2 + m · n). A b c d m B n a D c Proof. Drop a perpendicular from A to BC, hitting BC atE, and let h = AE and k = DE, as in the figure below. A c b h m a B d E n k D c From Pythagoras' Theorem, we have h2 + k2 = d2, so that and b2 = (n + k) 2 + h 2 = n 2 + 2nk + k 2 + h 2 = n 2 + 2nk + d 2. Multiplying the equation for c 2 by n and the equation for b2 by m, we have = n · m 2  2mnk + n · d 2, m · b2 = m · n 2 + 2mnk + m · d2 , n · c2 so that and since m n · c2 +n = +m · b2 = m · n( m + n) + (m + n) · d2 , a, then c2 · n + b2 · m =a· (d 2 + m · n). 0 EXAMPLES 71 Angle Bisector Theorem Theorem 3.4.7. (The Angle Bisector Theorem) Let D be a point on side BC of triangle ABC. = DB I DC. (2) If AD is the external bisector of LB AC, then AB I AC = DB I DC. (I) If AD is the internal bisector of LB AC, then AB I AC ~/ B C D (ii) (i) Proof. (1) Let l be a line parallel to AD through C meeting AB at E. Then LBAD:=LBEC and LCAD:=LACE, and so LAEC = LACE; that is, L.AC E is isosceles. Thus, AC=AE, and, again using the fact that CE II DA, L.ABD "' L.EBC. Hence, DB DC AB AE AB AC" The proof of (2) using similarity is left as an exercise. D Example 3.4.8. Use Stewart's Theorem to prove that in L.ABC, internal angle bisector at LA, then if AD= fA is the 72 SIMILARITY where a = BC, b = AC, and c = AB, as in the figure. A X X c b m B n D a c Solution. From the Internal Angle Bisector Theorem, we have m n so that and since a DB DC AB AC c b' c· n m= b, = m + n, then c· n a=m+n=+n b and That is, ab n=b+c and ac m= b+c" From Stewart's Theorem, we have that is, c b ) ( 2 a2bc ) abc ( b+c+b+c =a fA+(b+c)2 . Therefore, that is, D 73 EXAMPLES Corollary 3.4.9. In L.ABC, iff A, f B· and f c denote the lengths of the internal angle bisectors at LA, LB, and LC, respectively, then fi = be [ 1  f~ ac [1 = C: c) (a : r] ' 2 ] , c rJ. !8 = ab [1  (a : b Exercise 3.4.10. In 6ABC, let AD, BE, and CF be the angle bisectors of LBAC, LABC, and LACE, respectively. Show that if LBAC < LABC <LACE, then IADI > IBEI > ICFI. Medians In Chapter 2, we used area to show that the medians of a triangle trisect each other and are concurrent. Here is how similar triangles can be used to prove the same thing: Example 3.4.11. Let BE and C F be two medians of a triangle meeting each other at G. Show that EG FG 1 GB GC 2 and deduce that all three medians are concurrent at G. Solution. In the figure below, A B A B 74 SIMILARITY by the sAs similarity condition we have b.ABC rv b.AF E, with proportionality constant 1/2. This means that L.ABC L.AF E, from which it follows that FE is parallel to BC with FE = BC /2. This means that b.BCG rv b.EFG with a proportionality constant of 1/2. Hence, = ~GB GE= whichmeansthatEG/GB and GF = ~GC, = FG/GC = 1/2. Applying the same reasoning, medians BE and AD intersect at a point H for which EH/HB = HD/AH = 1/2. But this means that EG EH 1 GB HE 2' and since both G and Hare between Band E, we must have H that the three medians are concurrent at G. = G. This shows D We will end this section with a theorem that is sometimes useful in construction problems. Theorem 3.4.12. In b.ABC, suppose that D and E are points on AB and AC, respectively, such that DE II BC. Let P be a point on BC between Band C, and let Q be a point on DE between D and E. Then A, P, and Q are collinear if and only if BP DQ PC QE. The figures below illustrate the three cases that can arise. A A n~E Bz==]C p DV~E Q Proof. (i) Suppose A, P, and Q are collinear. Since DE is parallel to BC, we have b.APB rv b.AQD and b.APC rv b.AQE, CONSTRUCTION PROBLEMS 75 and so BP DQ  AP and AQ AP AQ PC QE' from which it follows that BP DQ PC QE' BP PC DQ QE' or, equivalently, that (ii) Conversely, suppose that BP DQ PC QE' Draw the line AP meeting DE at Q'. We will show that Q = Q'. Since A, P, and Q' are collinear, the first part of the proof shows that We are given that so that from which it follows that Q' BP PC DQ' Q'E' BP PC DQ QE' DQ' Q'E DQ QE' = Q, and this completes the proof. 0 3.5 Construction Problems If you were asked to bisect a given line segment, you would probably construct the right bisector. How would you solve the following problem? Example 3.5.1. To divide a given line segment into three equal parts. Many construction problems involve similarity. The idea is to create a figure similar to the desired one and then, by means of parallel lines or some other device, transform the similar figure into the desired figure. 76 SIMILARITY Creating a similar figure effectively removes size restrictions, and this can make a difficult problem seem almost trivial. Without size restrictions, Example 3.5 .1 becomes: To construct any line segment that is divided into three equal parts. This is an easy task: using any line, fix the compass at any radius, and strike off three segments of equal length AB, BC, and CD. Then AD is a line segment that has been divided into three equal parts. What follows shows how this can be used to solve the original problem. Solution. We are given the line segment PQ, which is to be divided into three equal parts. First, draw a ray from P making an angle with PQ. With the compass set at a convenient radius, strike off congruent segments P R', R' S', and S' Q' along the ray. p s R Q Join Q' and Q. Through R' and S', draw lines parallel to Q'Q that meet PQ at R and S. Since parallel projections preserve ratios, we have RS R'S' SQ = S'Q' = l, and so RS = SQ. Similarly, P R = RS. D The preceding construction can be modified to divide a line segment into given proportions. That is, it can be used to solve the following: given AB and line segments of length p and q, construct the point C between A and B so that AC CB p q CONSTRUCTION PROBLEMS 77 q p p A c B There is frequently more than one way to use similarity to solve a construction problem. Example 3.5.2. Given two parallel lines l and m and given points A on l, B on m, and P between l and m, construct a line n through P that meets l at C and m at D so that AC = ~BD, as in figure (a) below. Q A ',C n'' p ~p m ' ' ' ,D B (a) Q m D B (b) (c) Solution. (1) As in figure (b), draw the segment AB and extend it beyond A to Q so that BA = AQ. Draw the line PQ, meeting l at C and mat D, and then by similar triangles, AC = BD /2. (2) As in figure (c), on line l, strike off a segment AE. On line m, strike off segments BF and FG of the same length as AE. Let Q be the point where the lines AB and EG meet. Draw the line PQ, meeting l at C and mat D, and then by similar triangles, AC = BD /2. D 78 SIMILARITY Example 3.5.3. Given a point P inside an angle A, construct the two circles passing through P that are tangent to the arms of the angle. Solution. In the diagram, several construction arcs have been omitted for clarity. Analysis Figure: Construction: (1) Construct the angle bisector AB of LA. (2) Choose a point 0 1 on AB, and drop the perpendicular O' Q' to one of the arms of the angle. (3) Draw the circle C(O', O'Q'). This is the dotted circle in the diagram and it is tangent to both arms of the angle because 0' is on the angle bisector. (4) Draw the ray AP cutting the circle at P' and P". (5) Through P, construct the line PQ parallel to P'Q', with Q on AQ'. (6) Through Q, construct the line perpendicular to AQ meeting AO' at 0. (7) Draw the circle C(O, OQ). This is one of the desired circles. (8) Through P, construct the line PQ" parallel to P"Q', with Q" on AQ' (not shown). (9) Through Q", construct the line perpendicular to AQ" meeting AO' at 0" (not shown). ( 10) Draw the circle C(0", 0 11 Q") (not shown). This is the second desired circle. CONSTRUCTION PROBLEMS 79 Justification: To show that C(O, OQ) is one of the desired circles, we know that it is tangent to both arms of the angle because 0 is on the angle bisector and the radius OQ is perpendicular to an arm of the angle. It remains to show that the circle passes through P; that is, that OP = OQ, the radius of the circle. Since O'Q' Since Q' P' I OQ, it follows that OQ O'Q' AQ AQ'. QP Q'P' AQ AQ' OQ O'Q' QP Q'P'. II QP, it follows that and, consequently, that Again, since Q' P' I QP, it follows that LAQ' P' = LAQP, and so LO'Q'P' = LOQP. Then, by the sAs criteria, 60' Q' P' "' L:.OQ P, and so OP O'P' Thus OP OQ O'Q' OQ O'P'. = OQ, and this completes the proof. D Remark. Example 3.5.3 uses the idea that given a point A and a figure F, we can construct a similar figure g by constructing all points Q such that AQ = k · AP for points P in F, where k is a fixed nonzero constant. In the example, we constructed a circle centered at a point 0' and then constructed a magnified version of this circle centered at point 0. The transformation that maps each point P to the corresponding point Q so that AQ = k · AP is called a homothety and is denoted H(A, k). The number k is the magnification constant. The figure on the following page illustrates the effect of H(A,3). 80 SIMILARITY Q Q Here are two problems that are solved by translations rather than a homothety. The strategy is very similar: we construct a circle that fulfills part of the criteria and then use a translation to move it to the correct place. Example 3.5.4. Given a line l and points P and Q such that P and Q are on the same side of l and such that PQ _l_ l, construct the circle tangent to l that passes through P and Q. Solution. In the figure, we have omitted construction lines for the standard constructions (like dropping a perpendicular from a point to a line). Analysis Figure: m Construction: (1) Construct the right bisector m of PQ. (2) Choose any point 0' on m and drop the perpendicular O'T' from 0' to l. (3) ConstructC(O',r) wherer = O'T'. NotethatC(O',r) istangenttol. (4) Construct a line through P perpendicular to PQ that meets C(0', r) at P'. (5) With center 0' and radius P P', draw an arc cutting m at 0. (6) Draw C(0, r). This is the desired circle. CONSTRUCTION PROBLEMS 81 Justification: P P' 0' 0 is a parallelogram because P P' is congruent to and parallel to 00'. Then OP = 0' P' = r, showing that Pis on the circle C(O, r). Since 0 is on the right bisector of PQ, it follows that OQ = OP = r, showing that Q is on C(O, r). Finally, since m and l are parallel, the perpendicular distance from 0 to l is the same as the perpendicular distance from 0' to l, and it follows that C ( 0, r) is tangent to l. D Example 3.5.5. Given two disjoint circles of radii R and r, construct the four tangents to the circle. The diagram on the right illustrates that two nonoverlapping circles have four common tangent lines. The problem is to construct those lines. We will show how to construct the "external" tangents and leave the construction of the "internal" ones as an exercise. Solution. Analysis Figure: Let us suppose that the radii of the circles are R and r, as shown on the right. If we draw a line m through the center Q of the smaller circle parallel to a tangent line l as shown, then its distance from the center 0 of the larger circle will be R  r, and it will be tangent to C(O, R r). Construction: Here is the stepbystep construction. A 82 SIMILARITY (1) Draw a radius OA of the larger circle. With the compass, cut off a point B so that AB = r . Then OB = R r. Draw the circle C(O, OB). (2) Through the point Q, draw the tangents QS and QT to C( 0 , OB). (3) Draw the rays OS and OT, cutting the large circle at P and W , respectively. (4) Through P, draw a line parallel to QS. Through W , draw a line parallel to QT. These are the desired tangent lines. Justification: Let QU be parallel to SP so that QSPU is a rectangle. Then QU = PS = r where r is the radius of the smaller circle. Since PU ..l QU, then PU is tangent to the smaller circle. Since OP is a radius of the larger circle and since PU ..l OP, then PUis tangent to the larger circle. Let QV be parallel to TW so that QTWV is a rectangle. Reasoning as before, WV is tangent to both circles. D 3.6 The Power of a Point The following problem is taken from AHA! Insight, a delightful book written by Martin Gardner. Gardner attributes the problem to Henry Wadsworth Longfellow. A lily pad floats on the suiface of a pond as far as possible from where its root is attached to the bottom. If it is pulled out of the water vertically, until its stem is taut, it can be lifted I 0 em out of the water. The stem enters the water at a point 50 em from where the lily pad was originally floating. What is the depth of the pond ? ~ !Ocm 50 em Most people attack this problem by using Pythagoras' Theorem together with the fact that a chord is perpendicular to the radius that bisects it. At the end of this section we will see that there is a much more elegant approach. We begin with a valuable fact about intersecting secants and chords. THE POWER OF A POINT 83 Theorem 3.6.1. Let w be a circle, let P be any point in the plane, and let l and m be two lines through P meeting the circle at A and Band C and D, respectively. Then PA· PB =PC ·PD. Proof. If P is outside the circle, suppose that A is between P and B and that C is between P and D. Insert the line segments AD and BC, and consider triangles PADandPCB. m (J) A p By Thales' Theorem, L.PDA =L.PBC and L.P is common, so by the AA similarity criteria, L:.P AD "' L:.PC B. Consequently, PA PC and so PA·PB PD PB' = PC·PD. If P is inside the circle, again insert line segments AD and BC. Then, triangles PDA and PBC are similar, and so we again have PA · PB =PC· PD. D When P is outside the circle, an interesting thing happens if we swing the secant line PC D to a tangent position, as in the figure on the following page. In this case, the points C and D approach each other and coalesce at the point T, and so the product PC · P D approaches PT 2 . 84 SIMILARITY m l As a consequence: Theorem 3.6.2. Let P be a point in the plane outside a circle w. Let PT be tangent to the circle at T, and let l be a line through P meeting the circle at A and B. Then PT 2 =PA·PB. Proof. Here is a proof that does not involve limits. We may assume that A is between P and B. Insert the segments T A and T B, and we have the figure below. p A consequence of Thales' Theorem is that LPT A and LP BT are equal in size. It follows that Lo.PT A "' Lo.P BT, and so PT/PB = PA/PT. It follows immediately that PT 2 = P A· P B. D THE POWER OF A POINT If P is inside the circle, then, of course, no tangent to the circle passes through P. However, there is a result that looks somewhat like the previous theorem when P is inside the circle: 85 s Theorem 3.6.3. Let P be a point in the plane inside a circle w. LetTS be a chord whose midpoint is P and let AB be any other chord containing P. Then PT 2 =PA·PB. Now let l be a line, and assign a direction to the line. For two points A and Bon the line, with A i B, let ABbe the distance between A and B. The directed distance or signed distance from A to B, denoted AB, is defined as follows: AB= { AB 0 AB if A is before Bin the direction along l, if A= B, if B is before A in the direction along l. B A~ + AB=2 AB=2 BA=2 By using directed distances, the theorems above can be combined and extended as follows: Theorem 3.6.4. Let P be any point in the plane, let w be a given circle, and let l be a line through P meeting the circle at A and B. Then the value of the product P A · P B is independent of the line l, and ( 1) P A · P B > 0 if P is outside the circle, (2) P A · P B = 0 if Pis on the circle, (3) P A · P B < 0 if P is inside the circle. The value P A · P B is called the power of the point P with respect to the circle w. Suppose we are given a circle w with center 0 and radius r, and suppose that we are given a point P and that we know the distance 0 P. It is fairly obvious how we could experimentally determine the power of P with respect tow: draw any line through P meeting the circle at A and B, and measure P A and P B. Of course, if one chooses a line passing through 0, then P A and P Bare readily determined without recourse to measurement (see the figure on the following page). 86 SIMILARITY (l} (l} A PA = OPr PA PB=OP+r PB= r+OP = r OP By doing this, we find: Corollary 3.6.5. The power of a point P with respect to a circle with center 0 and radius r is OP 2  r 2 • The Lily Pad Problem The figure below illustrates the geometry of the lily pad problem. The depth of the pond is OP = x, and the length of the stem of the lily pad is the radius of a circle centered at 0. We know now that P A · P B = PT 2 , so referring to the diagram, (2x + 10)(10) and therefore the pond is 120 em deep. B X x+ 10 A = 50 2 , SOLUTIONS TO THE EXERCISES 87 Example 3.6.6. A person whose eyes are exactly 1 km above sea level looks towards the horizon at sea. The point on the horizon is 113 kmfrom the observer's eyes. Use this to estimate the diameter of the earth. Solution. The line of sight to the horizon is tangent to the earth. If x is the diameter of the earth, we have so x = 113 2  1 = 12768 km. 0 3.7 Solutions to the Exercises Solution to Exercise 3.3.2 There are many examples, as the following figures indicate: A A D D B B Solution to Exercise 3.3.7 We are given that AB DE BC AC = EF = DF = k, and so we have to prove that the angles are equal. 88 SIMILARITY D A c B Q E F We may assume that k < 1, that is, that AB <DE. Let P be the point on ED such that EP =EA. Draw PQ parallel to DF with Q on EF. By Theorem 3.2.4, 6EPQ rv 6EDF, and so EQ _ EP _ AB _ k _ BC EF  ED  ED  EF. This implies that EQ = BC. Similarly, PQ = AC, so 6EPQ =6BAC, and it follows that LDEF Since PQ II = LPEQ = LABC. D F we also have LEDF = LEPQ = LBAC and LEFD = LEQP = LBCA, and this completes the proof. Solution to Exercise 3.4.3 Given triangle ABC with AB 2 = BC 2 + CA 2 , let PQR be a triangle with a right angle at R, such that QR = BC and RP =CA. Then Hence, PQ = AB, so ABC and PQR are congruent triangles by SSS. It follows that LC is a right angle. Solution to Exercise 3.4.10 Let (3 = LCBE and let 1 =