Question

Line $l$ is the bisector of an angle $\angle A$ and $\angle B$ is any point on I. BP and BQ are perpendicular from B to the arms of $\angle A$ . Show that

$\left(i\right)\Delta APB\cong \Delta AQB$

$\left(ii\right)BP=BQ$ or $B$ is equidistant from the arms of $\angle A$.

Answer 1

Given:$l$ is the bisector of $\angle A$

So, $\angle PAB=\angle QAB$ .......$\left(1\right)$

$BP$ and $BQ$ are perpendiculars from $B$,

So,$\angle APB=\angle AQB={90}^{\circ }$ .......$\left(2\right)$

In $△APB$ and $△AQB,$

$\angle APB=\angle AQB$ from $\left(2\right)$

$\angle PAB=\angle QAB$ from $\left(1\right)$

$AB=AB$(common)

$\therefore △APB\cong △AQB$ by AAS congruence rule

$\therefore BP=BQ$ by CPCT

Hence proved.