proof of AAA (hyperbolic)

Following is a proof that AAA holds in hyperbolic geometry.

Proof.

Suppose that we have two triangles $\mathrm{△}ABC$ and $\mathrm{△}DEF$ 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. $\mathrm{\ell }(AC)=\mathrm{\ell }(DF)$

Before investigating the cases, $\mathrm{△}DEF$ will be placed on $\mathrm{△}ABC$ 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, $\mathrm{△}BEG$ 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, $\mathrm{△}BCE$ 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 $\mathrm{△}ABC$ and $\mathrm{△}DEF$ are congruent.

∎