MA341 071125 Notes on Chapter 6 This handout gives a summary of topics and definitions in Chapter 6. 1 Definitions. In the following, we assume that we have a Protractor Geometry (which means we have a Metric Geometry with Pasch Postulate and a protractor for all angles. We are about to add one more axiom and define the Neutral Geometry. 1. Comparisons of segments and angles. Given segments AB and CD, wesaythatab > CD if AB > CD. If AB = CD then we say AB = CD. Given angles ABC, DEF we say that ABC > DEF if m( ABC) > m( DEF) andifm( ABC) = m( DEF) thenwesay ABC = DEF. 2. Comparison of triangles. Given a triangle, ABC we shall find it convenient to abbreviate ABC to B and similarly for each of the three vertices. Given triangles ABC, DEF we say that they are congruent if there is a bijection F from S = {A, B, C} to T = {D, E, F} such that for each X S, wehave X = f(x) andalsoforx Y S, wehave XY = f(x)f(y ). Note that thus the notion of congruence involves checking six corresponding entities to be congruent (the three angles and the three sides). We shall write ABC = f(a)f(b)f(c). Warning. We shall not use the congruence symbol, unless we have the order of vertices arranged so that the corresponding entities are congruent. We will express the congruence by general words triangles ABC and DEF are congruent if we have not specified the correspondence between the vertices. Example. Thus, while it is true that triangles ABC and ACB are always congruent (when we match all vertices to themselves,) we can say ABC = ACB only if AB = AC and m( B) =m( C), i.e. the triangle is isoceles with AB = AC. 2 1 Preliminary Version.November 30, 2007 2 The theorem which says that AB = AC implies ABC = ACB is known as pons asimorum. It is amusing to look up what this means and why. This Theorem has a very easy proof due to Pappus, which is valid in a Neutral geometry. 1
3. The SAS axiom. In general, congruence of two triangles involves checking six congruences. In the Euclidean Geometry, it can be shown that a well chosen set of three of the six quantities is enough to deduce the congruence. We define: The SAS Axiom. Suppose that ABC, DEF are triangles. If AB = DE, B = E and BC = EF, then ABC = DEF. 4. Neutral (or Absolute) Geometry. A Protractor Geometry satisfying SAS axiom is said to be a Neutral Geometry. Both Euclidean and Poincare Geometries are Neutral. We shall deduce several congruence theorems and familiar facts assuming the Neutral Geometry hypothesis. It is tempting to think that a neutral Geometry shall have all the usual facts of Euclidean Geometry as true theorems, but that is not the case unless and until the so-called Parallel Postulate is added in. Thus we don t have the existence of a unique line M through a point P outside a line L so that L is parallel to M. We don t have the familiar theorem that the sum of the measures of the three angles of a triangle is a constant 180. Many theorems about angles between parallel lines and their transversals are also unavailable. Theorems in a Neutral Geometry. 1. Pons Asinorum. In a triangle ABC, ifab = AC then m( B) = m( C). Idea of proof. Argue ABC = ACB using SAS where the corresponding quantities are: AB = AC, m( A) =m( A),AC = AB. It follows that m( B) = m( C). 2. ASA Theorem. Given triangles ABC, DEF, if A = D, AB = DE and B = E then ABC = DEF. Idea of Proof. If AC = DF then we are done by SAS. Otherwise, assume without loss of generality, that AC < DF. By segment construction, there is a point G in DF with AC = DG. By SAS, ABC = DEG. 2
Then G is seen to be in the interior of DEF and hence DEG < DEF. But this is a contradiction since DEG = ABC = DEF. Here the first congruence follows from the proven congruence of triangles, while the second one is from hypothesis. 3. ASA implies SAS. It is possible to deduce the SAS statement as a theorem, if we assume the ASA statement as a hypothesis. Idea of the proof. Thus, assume we have triangles ABC, DEF with AB = DE, B = E,BC = EF. We wish to prove ABC = DEF using ASA as a known thoerem. If C = F, then we are done by ASA. We may assume without loss of generality that F < C and construct aray CD such that m( BCD) =m( F )andd is in the same side as A of the line BC. From hypothesis, the ray CD meets the crossbar AB, sayatsomepoint G, sothatwehavea G B. By ASA, we get GBC = DEF. But then, DE = GB < AB by betweenness and this contradicts the hypothesis AB = DE. 4. Converse of Pons Asinorum. In a triangle ABC, ifm( B) = m( C), then AB = AC. Idea of Proof. As before, argue that ABC = ACB by comparing the following ASA elements: B = C, BC = CB, C = B. The result follows from the congruence of triangles. 5. SSS Theorem. Given triangles ABC, DEF, if AB = DE, BC = EF,AC = DF then ABC = DEF. Idea of Proof. Choose a point Q on the opposite side of A from the line BC such that QB = DE and QBC = E. Then by SAS, deduce that DEF = QBC. Then it is enough to prove ABC = QBC. The segment AQ must then meet the line BC. 3
Let the intersection point be named G and list the five cases for its position on the line. In each case, by repeated application of Pons Asinorum, deduce that A = Q and then apply SAS axiom using the known equalities: to finish of the congruence. AB = DE = QB, AC = DF = QC 6. Relation between SSS and SAS. Unlike the clean equivalence of ASA and SAS, it is not known if SSS alone is equivalent to SAS. While SAS implies SSS, the converse can only be proved, for now, with a few extra assumptions. No counterexamples are known where SSS holds but SAS does not. Thus, this is an outstanding problem. 7. Perpendiculars. Given a line L and a point P outside it, there is a line M perpendicular to L containing P. Recall that this was easy from the protractor axiom when P was on L. Later we shall show the line M to be unique for a given P and L. Idea of Proof. Choose two points A, B on L and construct a point Q on the opposite side of L with respect to P such that m( QAB) = m( PAB)andPA = QA. The segment PQ intersects L at some point M. First suppose A M. Then we use SAS to prove that PAM = QAM. from this, we deduce that AMP = AMQ and since they form a linear pair, each has measure 90 as needed. In case A = M we get the two congruent angles PAB and QAB to form a linear pair and hence each is 90 as needed. This perpendicular will turn out to be unique after the Exterior Angle Theorem below. Note that the perpendicular line does meet the line L. 8. Exterior Angle Theorem. Given a triangle ABC choose a point P such that B C P holds. Then the angle ACP forms a linear pair with C with respect to the line BC and we call it the linear mate of C along the line BC. There is a similar linear mate with respect to the line AC. Moreover, these two mates are congruent, since they form a vertical pair. The Exterior Angle Thoeorem states that the measure of a linear mate of C is bigger than the measures of either of the other two angles A, B. These two angles are called the remote interior angles of the mate. 4
Idea of Proof. 3 To prove the result for the mate of C along BC set M to be the midpoint of AC and choose a point E such that M is the midpoint of BE. Then we can argue that AMB = CME. This gives m( A) = m( ECM) and the angle m( ECM) isseentobesmallerthanour exterior angle ACP. 9. Uses of the Exterior Angle Theorem. Let L be a line and P a point outside it. Given points A B C on L, we see that for the triangle PAB the exterior angle PBC at B has bigger measure than that of the remote interior angle m( PAB). Fix the point B and for any point X B on L define a function ψ(x) =m( PXB). If we let X vary over the interior of the ray CB then we see that this function ψ(x) steadily decreases as the distance XB increases. Moreover, all the values of ψ(x) asx varies over the interior of CB are less than m( PBC). In contrast, if we take X to be in the interior of the ray BC, theneach value ψ(x) islessthanm( PBA). Again, ψ(x) decreases as XB increases. It follows that there is a unique B such that PB is perpendicular to L. We list, without proof, a few more applications. SAA Theorem. This is a generalization of the ASA theorem, where we allow the two angles to be any two of the three angles, not necessarily the ones at either end of the side. In Euclidean Geometry where the sum of all three angle measures is 180, this trivially equivalent to ASA. In general Neutral geometry, we need a construction. Comparison of Sides and Angles. This is a natural supplement to the result about isoceles triangles. Consider a triangle ABC. Then AB > AC iff C > B. 3 If unique parallel lines were to exist, then the measure of the exterior mate of C equals the sum of the measures of the remote interior angles, namely m( A)+m( B). However, this is not necessarily true for a Neutral Geometry, hence we have this weaker statement. 5
Triangle Inequality. Triangle inequality holds in a Neutral Geometry. Idea of Proof. Given a triangle ABC, wechoosed with D B C such that DC = AB+BC and deduce that m( D) <m( DAC). It follows that the opposite sides AC and DC satisfy AC < DC = AB + BC. This proves the triangle inequality. Distance from a point to various points of a chosen line. Let L be a line and P a point outside L. ChooseapointB on L such that the line PB is perpendicular to L. Let f be a ruler on L such that f(b) =0. LetPB = d. Define the function θ from L to R such that θ(x) =PX for each point X on L. The line is then a union of two rays R + = {X f(x) 0} and R = {X f(x) 0}. The only common point of the two rays is B. It can be shown that θ is an injective function on each of these rays with range [d, ). 4 Moreover, for any two points X 1,X 2 on L, wehave f(x 1 ) > f(x 2 ) iff θ(x 1 ) >θ(x 2 ). 4 Is it surjective? To be continued. 6