Question

In the given figure, 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 $\angle A.$ Show that $B$ is equidistant from the arms of $\angle A.$

Answer 1

In $△$s $APB$ and $ABQ$, we have
$\angle APB=\angle AQB$ (Each ${90}^{\circ }$)
$\angle PAB=\angle QAB$ ($AB$ bisect $\angle PAQ$)
$AB=BA$ (common)
Therefore, $△APB\cong △ABQ$ (AAS)
$\Rightarrow$ $BP=BQ$ (cpct)
Hence, $B$ is equidistant from the anus of $\angle A$.