Question

Line $l$ is the bisector of an angle $\angle A$ and $B$ is any point on $l$. $BP$ and $BQ$ are perpendiculars from $B$ to the arms of $\angle A$. Show that:
(i) $△ABP\cong △AQB$
(ii) $BP=BQ$ or $B$ is equidistant from the arms of $\angle A$.

Answer 1

In $△APB$ and $△AQB$

$\angle APB=\angle AQB$ (Each ${90}^o$)

$\angle PAB=\angle QAB$ ($l$ is the angle bisector of $\angle A$)

$AB=AB$ (Common)

$\therefore △APB\cong △AQB$ (By AAS congruence rule)

$\therefore BP=BQ$ (By CPCT)

It can be said that $B$ is equidistant from the arms of $\angle A$.