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

(N/A) Given: Lines $l$ and $m$ intersect at point $A$. $P$ is a point such that $PB \perp l$ and $PC \perp m$,with $PB = PC$.
To prove: Line $AP$ bisects the angle between lines $l$ and $m$,i.e.,$\angle PAB = \angle PAC$.
Proof: Consider $\Delta PAB$ and $\Delta PAC$.
In these two triangles:
$1$. $PB = PC$ (Given,as $P$ is equidistant from the lines)
$2$. $\angle PBA = \angle PCA = 90^o$ (Given,as $PB \perp l$ and $PC \perp m$)
$3$. $PA = PA$ (Common side)
Therefore,by the $RHS$ congruence rule,$\Delta PAB \cong \Delta PAC$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$\angle PAB = \angle PAC$.
This shows that the line $AP$ bisects the angle between lines $l$ and $m$.