Question

$P$ is a point equidistant from two lines $l$ and $m$ intersecting at point $A$. Show that the line $AP$ bisects that angle between them

Answer 1

You are given that lines $l$ and $m$ intersect each other at $A$.

Let $PB$ is perpendicular on $l$ and

$PC\perp m$. It is given that $PB=PC$.

You need to show that $\angle PAB=\angle PAC$.

Let us consider $△PAB$ and $△PAC$. In these two triangles,

$PB=PC$ (Given)

$\angle PBA=\angle PCA={90}^{\circ }$ (Given)

$PA=PA$ (Common side)

So, $△PAB\cong △PAC$ (RHS rule)

So, $\angle PAB=\angle PAC\left(CPCT\right)$