Two lines $l$ and $m$ intersect at the point $O$ and $P$ is a point on a line $n$ passing through the point $O$ such that $P$ is equidistant from $l$ and $m$. Prove that $n$ is the bisector of the angle formed by $l$ and $m$.

(N/A) Given: Lines $l$ and $m$ intersect at point $O$. $P$ is a point on line $n$ passing through $O$ such that $PQ \perp l$ and $PR \perp m$,where $PQ = PR$.
To prove: $n$ is the bisector of $\angle QOR$.
Proof: In $\triangle OQP$ and $\triangle ORP$,we have:
$\angle OQP = \angle ORP = 90^{\circ}$ (Given that $P$ is equidistant from $l$ and $m$)
$OP = OP$ (Common side)
$PQ = PR$ (Given)
By $RHS$ (Right angle-Hypotenuse-Side) criterion of congruence,we have:
$\triangle OQP \cong \triangle ORP$
Therefore,$\angle QOP = \angle ROP$ (by $CPCT$ - Corresponding Parts of Congruent Triangles).
Thus,$n$ is the bisector of $\angle QOR$. Hence proved.