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

$\left(i\right)△APB\cong △AQB$

$\left(ii\right)BP=BQ$ 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 $AAS$ congruency rule]

$\therefore BP=BQ$ [by $CPCT$]

Hence proved.