Problem — Congruence criterion with exterior angles

Text

Prove that two triangles are congruent if they have, respectively congruent, a side and the two exterior angles at the endpoints of that side.

Solution

Let $ABC$ and $A'B'C'$ be two triangles such that $AB \cong A'B',$ and let the exterior angles at vertices $A$ and $B$ be congruent to those at vertices $A'$ and $B'$ respectively.

An exterior angle is the supplement of the corresponding interior angle. Hence $\widehat{A} = 180^\circ - (\text{exterior angle at } A), \qquad \widehat{A'} = 180^\circ - (\text{exterior angle at } A').$ Since the exterior angles at $A$ and $A'$ are congruent, so are their supplements, i.e. the interior angles: $\widehat{A} \cong \widehat{A'}.$ By the same reasoning at vertices $B$ and $B'$: $\widehat{B} \cong \widehat{B'}.$

The two triangles then have congruent the side $AB \cong A'B'$ and the two interior angles adjacent to it: $\widehat{A} \cong \widehat{A'}, \qquad \widehat{B} \cong \widehat{B'}.$ By the second congruence criterion (ASA) we conclude $\triangle ABC \cong \triangle A'B'C'. \qquad \blacksquare$