Problem — The exterior angle at the base of an isosceles triangle is obtuse

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Let $MNP$ be an isosceles triangle with base $MN$ (so that $PM \cong PN$). Extend the base $MN$ beyond the vertex $M$ to a point $S$. Prove that the exterior angle $\widehat{PMS}$, adjacent to the base angle $\widehat{PMN}$, is always obtuse.

Solution

Let $\varphi$ denote the measure of the apex angle $\widehat{MPN}$ and $\beta$ the measure of each base angle. Since the triangle is isosceles with base $MN$, the base angles are congruent: $\widehat{PMN} \cong \widehat{PNM} = \beta$.

From the sum of the interior angles of the triangle, $2\beta + \varphi = 180^\circ \quad\Longrightarrow\quad \beta = 90^\circ - \frac{\varphi}{2}.$ Since the triangle is non-degenerate, the apex angle satisfies $\varphi > 0^\circ$, hence $\beta = 90^\circ - \dfrac{\varphi}{2} < 90^\circ$: every base angle is acute.

The exterior angle $\widehat{PMS}$ is adjacent to the base angle $\widehat{PMN}$, therefore $\widehat{PMS} = 180^\circ - \beta = 180^\circ - \left(90^\circ - \frac{\varphi}{2}\right) = 90^\circ + \frac{\varphi}{2}.$ Since $\varphi > 0^\circ$, we have $\widehat{PMS} = 90^\circ + \dfrac{\varphi}{2} > 90^\circ$: the exterior angle at the base is therefore always obtuse. $\;\blacksquare$