Bisector of Apex of Isosceles Triangle is Perpendicular to Base

Theorem

Let $\triangle ABC$ be an isosceles triangle whose apex is $A$.

Let $AD$ be the bisector of $\angle BAC$ such that $AD$ intersects $BC$ at $D$.

Then $AD$ is perpendicular to $BC$.

Proof

By definition of isosceles triangle, $AB = AC$.

By definition of bisector, $\angle BAD = \angle CAD$.

By construction, $AD$ is common.

Thus by Triangle Side-Angle-Side Congruence, $\triangle ABD = \triangle ACD$.

Thus $\angle ADB = \angle ADC$.

By Two Angles on Straight Line make Two Right Angles, $\angle ADB + \angle ADC$ equals $2$ right angles.

Thus each of $\angle ADB$ and $\angle ADC$ are right angles.

The result follows by definition of perpendicular.

$\blacksquare$

Sources

• 1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.12$: Corollary $1 \ \text{(ii)}$