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.