PlanetMath: proof of AAA (hyperbolic)

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:

$\ell(AC)>\ell(DF)$
$\ell(AC)=\ell(DF)$
$\ell(AC)<\ell(DF)$

Before investigating the cases, $\triangle DEF$ will be placed on $\triangle ABC$ so that the following are true:

and correspond
, , and are collinear
, , and are collinear

Now let us investigate each case.

Case 1: Let denote the intersection of $\overline{BC}$ and $\overline{EF}$

$\begin{pspicture}(-2,-2)(2,2.5) \psline(-1.5,-1.5)(0,2) \psline(-1.5,-1.5)(1.17,... ....4,-0.8){$F$} \rput[1](1.5,-1.5){$C$} \rput[a](0.407,-1.35){$G$} \end{pspicture}$

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:

$\begin{pspicture}(-2,-2)(2,2.5) \psline(-1.5,-1.5)(0,2) \psline(-1.5,-1.5)(1.35,... ...5,-0.8){$B$} \rput[r](-1.55,-1.5){$E$} \rput[l](1.4,-1.2){$C=F$} \end{pspicture}$

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.

$\begin{pspicture}(-2,-2)(2,2.5) \psline(-1.5,-1.5)(0,2) \psline(-1.5,-1.5)(1.3,-... ...-1.55,-1.5){$E$} \rput[1](1.4,-0.8){$C$} \rput[1](1.5,-1.5){$F$} \end{pspicture}$

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, 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.

$\qedsymbol$