In class, we mentioned that (AAA) is a triangle congruence in a neutral geometry if and only if this neutral geometry is hyperbolic. Let's prove this result here. (a) First, we present a proof that, if a neutral geometry is hyperbolic, then (AAA) is a triangle congruence. More precisely, let AABC and ADEF be triangles in a hyperbolic geometry and suppose ZA ZD, ZB ZE and ZC ZF. We will prove that AABC = ADEF. Justify each line of the proof below. i. Assume for a contradiction that AABC # ADEF. Without loss of generality, we may assume that DF < AC. In this case, there exists a point A' with A* A' *C and A'C=DF. ii. There exists a point P with PE HAC,B and ZCA'P = LFDE. iii. There exists a point B' with A'P BC = {B'}. iv. Note that AA'B'C= ADEF. v. Note that 8(AABC) = 8(AAA'B') + 6(AAB'B') + 8(AA'B'C'). vi. It follows that 8(AABC) > $(AA'B'C'). vii. This is a contradiction. (b) Which line from the proof in (a) is false in Euclidean geometry? (c) Now, show that (AAA) is not a triangle congruence in Euclidean geometry. More precisely, given a triangle AABC in a Euclidean geometry, provide an algorithm to construct a triangle ADEF with D = A, E & AB. d(D, E) = 2d(A,B), F & A, ZFDE ZCAB, ZDEF = ZABC and ZDFE ZACB.