Question

$AD$ is perpendicular to the bisector of the $\angle B$ of $\Delta ABC$. $DE$ is drawn through $D$ and parallel to $BC$ to meet $AC$ at $E$. Prove that $AE=EC$.

Answer 1

$AD$ extended meet $BC$ at $F.$

$\angle ADB=\angle FDB={90}^{\circ }$

$\angle ABD=\angle FBD$ ($BD$ is angle bisector of $\angle B$)

$\therefore \angle BAD=\angle BFD$

$△ADB$ and $△FDB$ are congruent.

So, $AD=DF$ (BY CPCT)

$AD\parallel FC$ and $D$ is mid point of $AF.$

By mid point theorem,

$E$ is mid-point of $AC.$

Hence, $AE=EC$