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Homeproof of AAA (hyperbolic)
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proof of AAA (hyperbolic)
Following is a proof that AAA holds in hyperbolic geometry.
Proof.
Suppose that we have two triangles $\triangle ABC$ and $\triangle DEF$ 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 $\ell(AB)<\ell(DE)$, where $\ell$ is used to denote length. (Note that, if $\ell(AB)=\ell(DE)$, then the two triangles would be congruent by ASA.) Then there are three cases:
1. $\ell(AC)>\ell(DF)$
2. $\ell(AC)=\ell(DF)$
3. $\ell(AC)<\ell(DF)$
Before investigating the cases, $\triangle DEF$ will be placed on $\triangle ABC$ 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 $\angle ABC$ and $\angle CBE$ are supplementary. By hypothesis, $\angle ABC$ and $\angle DEF$ are congruent. Thus, $\angle CBE$ and $\angle DEF$ are supplementary. Therefore, $\triangle BEG$ contains two angles which are supplementary, a contradiction.
Case 2:
Note that $\angle ABC$ and $\angle CBE$ are supplementary. By hypothesis, $\angle ABC$ and $\angle DEF$ are congruent. Thus, $\angle CBE$ and $\angle DEF$ are supplementary. Therefore, $\triangle BCE$ 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 $\angle ABC$ and $\angle CBE$ are supplementary. By hypothesis, $\angle ABC$ and $\angle DEF$ are congruent. Thus, $\angle CBE$ and $\angle DEF$ are supplementary. Similarly, $\angle BCF$ and $\angle DFE$ 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 $\triangle ABC$ and $\triangle DEF$ are congruent.
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