Question

In the given figure, AE is the bisector of $\angle BAC$ and $AD\perp BC$

Show that $\angle DAE=\frac{1}{2}(\angle C-\angle B)$

Answer 1

AE is bisector of $\angle BAC$

$\therefore \angle CAE=\frac{\angle A}{2}$

$\angle A+\angle B+\angle C=180°$

$\angle A=180°-\angle B-\angle C$

$\frac{\angle A}{2}=90°-\frac{\angle B}{2}-\frac{\angle C}{2}$

AD is perpendicular to BC, $\therefore \angle CAD=90°-\angle C$

$\angle DAE=\angle CAE-\angle CAD$

$=90°-\frac{\angle B}{2}-\frac{\angle C}{2}-90°+\angle C$

$\angle DAE=\frac{1}{2}(\angle C-\angle B)$