proof of AAA (hyperbolic)
Following is a proof that AAA holds in hyperbolic geometry.
Proof.
Suppose that we have two triangles^{} $\mathrm{\u25b3}ABC$ and $\mathrm{\u25b3}DEF$ such that all three pairs of corresponding angles are congruent, but that the two triangles are not congruent. Without loss of generality, let us further assume that $$, where $\mathrm{\ell}$ is used to denote length. (Note that, if $\mathrm{\ell}(AB)=\mathrm{\ell}(DE)$, then the two triangles would be congruent by ASA.) Then there are three cases:

1.
$\mathrm{\ell}(AC)>\mathrm{\ell}(DF)$

2.
$\mathrm{\ell}(AC)=\mathrm{\ell}(DF)$

3.
$$
Before investigating the cases, $\mathrm{\u25b3}DEF$ will be placed on $\mathrm{\u25b3}ABC$ so that the following are true:

•
$A$ and $D$ correspond

•
$A$, $B$, and $E$ are collinear^{}

•
$A$, $C$, and $F$ are collinear
Now let us investigate each case.
Case 1: Let $G$ denote the intersection^{} of $\overline{BC}$ and $\overline{EF}$
Note that $\mathrm{\angle}ABC$ and $\mathrm{\angle}CBE$ are supplementary^{}. By hypothesis^{}, $\mathrm{\angle}ABC$ and $\mathrm{\angle}DEF$ are congruent. Thus, $\mathrm{\angle}CBE$ and $\mathrm{\angle}DEF$ are supplementary. Therefore, $\mathrm{\u25b3}BEG$ contains two angles which are supplementary, a contradiction^{}.
Case 2:
Note that $\mathrm{\angle}ABC$ and $\mathrm{\angle}CBE$ are supplementary. By hypothesis, $\mathrm{\angle}ABC$ and $\mathrm{\angle}DEF$ are congruent. Thus, $\mathrm{\angle}CBE$ and $\mathrm{\angle}DEF$ are supplementary. Therefore, $\mathrm{\u25b3}BCE$ contains two angles which are supplementary, a contradiction.
Case 3: This is the most interesting case, as it is the one that holds in Euclidean geometry^{}.
Note that $\mathrm{\angle}ABC$ and $\mathrm{\angle}CBE$ are supplementary. By hypothesis, $\mathrm{\angle}ABC$ and $\mathrm{\angle}DEF$ are congruent. Thus, $\mathrm{\angle}CBE$ and $\mathrm{\angle}DEF$ are supplementary. Similarly, $\mathrm{\angle}BCF$ and $\mathrm{\angle}DFE$ are supplementary. Thus, $BCFE$ is a quadrilateral^{} whose angle sum is exactly $2\pi $ radians, a contradiction.
Since none of the three cases is possible, it follows that $\mathrm{\u25b3}ABC$ and $\mathrm{\u25b3}DEF$ are congruent.
∎
Title  proof of AAA (hyperbolic) 

Canonical name  ProofOfAAAhyperbolic 
Date of creation  20130322 17:08:46 
Last modified on  20130322 17:08:46 
Owner  Wkbj79 (1863) 
Last modified by  Wkbj79 (1863) 
Numerical id  6 
Author  Wkbj79 (1863) 
Entry type  Proof 
Classification  msc 51M10 