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Chapter 3 Foundations of Geometry 2

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Chapter 3 Foundations of Geometry 2 Chapter 3 Foundations of Geometry 2 3.1 Triangles, Congruence Relations, SAS Hypothesis De¯nition 3.1 A triangle is the union of three segments ( called its side), whose end points (called its vertices) are taken, in pairs, from a set of three noncollinear points. Thus, if the vertices of a triangle are A; B; and C, then its sides are AB; BC; AC, and the triangle is then the set de¯ned by AB [ BC [ AC, denoted by ¢ABC. The angles of ¢ABC are \A ´ \BAC, \B ´ \ABC, and \C ´ \ACB. Equality and Congruence Equal(=): identically the same as. For examples, 1) Two points are equal, as in A = B, we mean they coincide. 2) AB = CD only if the set of points AB is the exact same set of points denoted by CD. Congruence(»=): is an equivalence relation. (to be de¯ned later). Congruence for Segments and angles 25 Yi Wang Chapter 3. Foundations of Geometry 2 26 AB »= XY i® AB = XY \ABC »= \XYZ i® m\ABC = m\XYZ Congruence for triangles Notation: correspondence between two triangles Given ¢ABC and ¢XYZ, write ABC $ XYZ to mean the correspondence between vertices, sides and angles in the order written. Note: There are possible six ways for one triangle to correspond to another. ABC $ XY Z ABC $ XZY ABC $ YXZ ABC $ Y ZX ABC $ ZXY ABC $ ZYX De¯nition 3.2 (Congruence for triangles) If, under some correspondence between the vertices of two triangles, corresponding sides and corresponding angles are congruent, the triangles are said to be congruent. Thus we write ¢ABC »= ¢XYZ whenever AB »= XY; BC »= YZ; AC »= XZ; \A »= \X; \B »= \Y; \C »= \Z Notation: CPCF: Corresponding parts of congruent ¯gures (are congruent). Properties of Congruence 1. reflexive law: ¢ABC »= ¢ABC 2. Symmetry Law: If ¢ABC »= ¢XYZ, then ¢XYZ »= ¢ABC 3. Transitive Law: If ¢ABC »= ¢XYZ, and ¢XYZ »= ¢UVW , then ¢ABC »= ¢UVW . Remark: Euclid did not use the word congruence. Euclid attributed to congruent triangles the property that on triangle could be placed precisely on top of another. Yi Wang Chapter 3. Foundations of Geometry 2 27 3.2 Taxicab Geometry: Geometry without SAS con- gruence Skip 3.3 SAS, ASA, SSS Congruence and Perpendicular Bi- sectors Question: Can we require fewer than six sets of congruent pairs to determine two triangles are congruent? The SAS Hypothesis Under the correspondence ABC $ XYZ, let two sides and the included angles of ¢ABC be congruent, respectively, to the corresponding two sides and the included angle of ¢XYZ. Note: This can not be established within the current set of axioms. (See Section 3.2. Axiom 3.3 (SAS Postulate) If the SAS Hypothesis holds for two triangles under some correspondence between their vertices, then the triangles are congruent. Theorem 3.4 (ASA Theorem) If, under some correspondence, two angles and the in- cluded side of one triangle are congruent to the corresponding angles and included side of another, the triangles are congruent under that correspondence. Proof: Outline of the proof: (Use SAS postulate). Given \A »= \X, AB »= XY , and \B »= \Y . If AC »= XZ, then the two triangles are congruent by SAS. But if not, then either AC > XZ or AC < XZ. 1) Assume AC > XZ, then we can ¯nd D on AC such that A ¡ D ¡ C and AD »= XZ by Yi Wang Chapter 3. Foundations of Geometry 2 28 the Segment Construction Theorem. This will leads to a contradiction. 2) One can argue similarly for the case AC < XZ. 2 Example 3.5 Given: \EBA »= \CBD; AB »= BC; \A »= \C. Prove: EB »= DB. Conclusions Justi¯cations (1) ¢ABE »= ¢CBD ASA (2) ) EB »= DB CPCF Example 3.6 Yi Wang Chapter 3. Foundations of Geometry 2 29 Given: M is the midpoint of CD and EF . Prove: \C = \D. Example 3.7 Prove that a line that bisects an angle also bisects any segment perpendicular to it that joins two points on the sides of that angle. Isosceles Triangle Theorem De¯nition 3.8 Isosceles triangle: a triangle having two sides congruent. Legs, base, base angles, vertex, vertex angle. Lemma 3.9 In ¢ABC, if AC »= BC, then \A »= \B. Proof: By the SAS postulate: ¢CAB »= ¢CBA, therefore, corresponding angles \CAB and \CBA are congruent, that is \A »= \B. (You could also construct an angle bisector from the vertex). 2 Theorem 3.10 A triangle is isosceles i® the base angles are congruent. Proof: Only need to show the converse. By the ASA theorem, ¢CAB »= \CBA. Therefore AC »= BC. 2 Example 3.11 Solve the following problem in proof writing. Yi Wang Chapter 3. Foundations of Geometry 2 30 Given: GJ = KM = 2;JK = HJ = HK = 10, with betweenness relations as evident from the ¯gure. Prove: HG = HM. Symmetry á! Lemma 3.12 If M is the midpoint of segment AB and line PM is perpendicular to AB, then PA = PB. Lemma 3.13 If PA = PB and M is the midpoint of segment AB, then line PM is per- pendicular to segment AB. Lemma 3.14 If PA = PB and M is the midpoint of segment AB, then line PM bisects \AP B. Perpendicular Bisectors, Locus De¯nition 3.15 The Perpendicular bisector of a segment AB to be the line that both bisects AB and is perpendicular to it. Locus: The locus of a point is the path or set of points that is determined by that point when it satis¯es certain given properties. Theorem 3.16 (Perpendicular Bisector Theorem) The set of all points (The locus of a point which is ) equidistant from two distinct points A and B is the perpendicular bisector of the segment AB. Yi Wang Chapter 3. Foundations of Geometry 2 31 Remark: To prove above theorem, actually one needs to prove two sets are equal. Namely, the Locus (set of points, in this theorem, it is the perpendicular bisector) is equal to the set of points which are equidistant from the two distinct points A and B. Proof: ()) Let ` be the perpendicular bisector of AB, and assume P 2 `. By Lemma 3.12, PA = PB. ((): Assume P is equidistant from A and B. Let M be the midpoint of AB, then by Lemma á! á! 3.13, PM?AB. (Next argue PM = `). But `?AB, there can be only one perpendicular to á! AB at M. Hence PM = `, and P 2 `. 2 Theorem 3.17 If, under some correspondence between their vertices, two triangles have the three sides of one congruent to the corresponding three sides of the other, then the triangles are congruent under that correspondence. Proof: class exercise. Note, there are three di®erent cases, but can be reduced to two cases. 2 Example 3.18 Name all congruent pairs of distinct segments and angles in the following ¯gure. Example 3.19 In the following ¯gure, we have a four-sided ¯gure with the congruent sides and right angles as marked. (a) Prove that AD »= CD. (b) If, in addition, BC »= CD, prove \D is a right angle. Yi Wang Chapter 3. Foundations of Geometry 2 32 Theorem 3.20 (Existence of perpendicular from an external point) Let the line ` and point A not on ` be given. Then there exists a unique line m perpendicular to ` passing through A. Proof: (Existence) By construction. Locate B and C on `. let \DBC »= \ABC; BD »= BA. It follows that B and C are both equidistant from A and D. (Why) Hence AD?BC by the perpendicular bisector theorem. (Uniqueness): Class discussion. 2 3.4 Exterior Angle Inequality Ã! De¯nition 3.21 Let ¢ABC be given, and suppose D is a point on BC such that the be- tweenness relation B ¡ C ¡ D holds. Then \ACD is called an exterior angle of the given triangle. The angles at A and B of ¢ABC are called opposite interior angles of \ACD. Yi Wang Chapter 3. Foundations of Geometry 2 33 Remark: 1) In Euclidean geometry: The measure of an exterior angle of any triangle equals the sum of the measures of the other two opposite interior angles. 2) in absolute geometry: the above relationship is not valid in general. De¯nition 3.22 Absolute geometry consists of part of Euclidean geometry that includes all the axioms except all references to parallel lines, or results therefrom. Absolute geometry provides foundations not only for Euclidean geometry, but also for non-Euclidean geometry. Exterior angles: example provided by Henri Poincar`e Poincar¶eModel for Absolute Geometry Yi Wang Chapter 3. Foundations of Geometry 2 34 1. C is a circle with center O; 2. All points inside C are \points" of this geometry. A point on or outside the circle is not a \point"; 3. \line": either an straight line through O, cut o® by C, or the arc of a circle that makes right angle with C and the arc lies inside C; 4. Betweeness: If on an arc of a circle point Q is between points P and R, then we de¯ne P ¡ Q ¡ R. And the de¯nitions for segments, rays, and angles follow this betweenness de¯nition. 5. for example, the \segment" AB shown in the following ¯gure. Yi Wang Chapter 3. Foundations of Geometry 2 35 6. Angle measure: the angle between two curves de¯ned as the angle measure of the angle formed by the tangents to the two curves at the point of intersection. 7. The geometry inside C satis¯es all the axioms for absolute geometry in plane. For example: two points determine a unique line. 8. Consider one of the triangles: (see the following ¯gure) Yi Wang Chapter 3.
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