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:
- $A$ 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.
