Question

Line $ l$ is the bisector of an angle $ A$ and $ B$ is any point on $ l$. $ BP$ and $ BQ$ are perpendiculars from $ B$ to the arms of $ A$ (see Fig.).

Show that:

(i) $ △APB\cong △AQB$
(ii) $ BP=BQ$ or $ B$ is equidistant from the arms of $ A$.

Answer 1

Step 1: To prove $△APB\cong △AQB$:

Given,

Line $l$ is the bisector of angle $A$

$B$ is a point on $l$.

$BP$ and $BQ$ are perpendiculars from $B$ to the arms of $A$.

In $△APB$ and $△AQB$,

$\Rightarrow$ $AB=AB$ $[\because$Common side$]$
$\Rightarrow$ $\angle BAQ=\angle PAB$ $[\because$Both are angles formed by bisecting $\angle A]$
$\Rightarrow$ $\angle BQA=\angle BPA$ $[\because$Both are right angles$]$

$\therefore △APB\cong △AQB$ $[$By AAS congruency$]$

Hence proved.

Step 2: To prove $BP=BQ$:

We know that, $△APB\cong △AQB$

Since $BP$ and $BQ$ are corresponding sides of congruent triangles

$\therefore BP=BQ$ or $B$ is equidistant from the arms of $A$.

Hence proved.