Question

Line $l$ is the bisector of an angle $\angle A$ and $\angle B$ is any point on $l$. $BP$ and $BQ$ are perpendicular from $B$ to the arms of $\angle A$ (see figure). Show that
(i) $\Delta APB=\Delta AQB$
(ii) $BP=BQ$ or $B$ is equidistant from the arms of $\angle A$.

Answer 1

(i) In $\Delta APB$ and $\Delta AQB$
$\angle BQA=\angle BPA={90}^0$ (given)
$AB=AB$ (common side)
and $\angle QAB=\angle PAB$
($\because l$ is the bisector of $\angle A$)
$\therefore \Delta APB\cong \Delta AQB$ (by AAS congruence rule)
(ii) $\because \Delta APB\cong \Delta AQB$
$BP=BQ$ (by c.p.c.t)
i.e. $B$ is equidistant from the arms of $\angle A$.