Problem — Congruence of isosceles triangles with equal apex angle

Text

Prove that two isosceles triangles are congruent if they have congruent apex angle and one congruent side. Discuss the case according to whether the congruent side is a leg or the base.

Solution

Let $ABC$ (with $AB \cong AC$, apex $A$) and $A'B'C'$ (with $A'B' \cong A'C'$, apex $A'$) be two isosceles triangles with congruent apex angles: $\widehat{A} \cong \widehat{A'}.$ The “congruent side” can be of two kinds.

Case 1: the congruent side is a leg. Suppose $AB \cong A'B'$. Since the triangles are isosceles, the other legs are congruent to these as well: $AC \cong AB \cong A'B' \cong A'C'.$ The two triangles then have two congruent sides ($AB \cong A'B'$, $AC \cong A'C'$) and the included angle congruent ($\widehat{A} \cong \widehat{A'}$). By the first criterion (SAS), $\triangle ABC \cong \triangle A'B'C'.$

Case 2: the congruent side is the base. Suppose $BC \cong B'C'$. In an isosceles triangle the base angles are congruent and equal $\widehat{B} = \widehat{C} = \frac{180^\circ - \widehat{A}}{2}, \qquad \widehat{B'} = \widehat{C'} = \frac{180^\circ - \widehat{A'}}{2}.$ Since $\widehat{A} \cong \widehat{A'}$, we have $\widehat{B} \cong \widehat{B'}$ and $\widehat{C} \cong \widehat{C'}$. The two triangles thus have congruent bases and congruent adjacent angles: by the second criterion (ASA), $\triangle ABC \cong \triangle A'B'C'.$

In both cases the triangles are congruent: the single congruent side, whether a leg or the base, together with the apex angle suffices. $\blacksquare$