The goal of this problem is a proof of the following:
Hyperbolic Angle-Angle-Angle (Hyperbolic AAA) Theorem.
Consider two triangles ∆A1B1C1 and ∆A2B2C2, with angle measures α1 at A1, β1 at B1, γ1 at C1 and α2 at A2, β2 at B2, γ2 at C2, and side lengths a1 = B1C1, b1 = A1C1, c1 = A1B1 and a2 = B2C2, b2 = A2C2, c2 = A2B2. If α1 = α2, β1 = β2, and γ1 = γ2, then ∆A1B1C1 ≅ ∆A2B2C2; i.e., a1 = a2, b1 = b2, and c1 = c2.
Step 1. Suppose c1 < c2. Explain why we may then construct a point D on A2B2 with A2D = c1 and a line segment A2E of length b1 with E on A2C2 or C2 on A2E. In particular, D ≠A2 and D ≠B2 since c1 < c2. Then explain why we may construct DE and why ∆A2DE ≅ ∆A1B1C1. Include a picture.
Step 2. Assuming still that c1 < c2, prove that b1 < b2 and hence that E is between A2 and C2 (and E ≠A2 and E ≠C2).
Hint. To prove it by contradiction, assume that b1 ≥ b2. Show that the measure of ∠DB2E is greater than or equal to β2 = β1, and apply the Hyperbolic Exterior Angle Theorem to ∆DB2E, ∠A2DE, and β1 = β2. Include a picture.
Step 3. Complete the proof. Include a picture.
Hint. Assuming still that c1 < c2, and hence that b1 < b2, show that the convex quadrilateral DB2C2E has an angle sum of 360°. Explain why this is a contradiction.