Question

$P$ is a point equidistant from two lines $l$ and $m$ intersecting at point $A$.Show that $△PAB\cong △PAC$

Answer 1

Given that lines $l$ and $m$ intersect each other at $A$

Let $PB\perp l,PC\perp m$

In $△PAB$ and $△PAC,$

$PB=PC$(given)

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

$PA=PA$(common)

So,$△PAB\cong △PAC$ by RHS rule.