Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

(N/A) Given: Two tangents $PR$ and $PQ$ are drawn from an external point $P$ to a circle with centre $O$.
To prove: The centre $O$ of the circle lies on the angle bisector of the angle formed by the two intersecting lines $PR$ and $PQ$.
Construction: Join $OR$ and $OQ$.
Proof:
In $\triangle PRO$ and $\triangle PQO$:
$1$. $\angle PRO = \angle PQO = 90^{\circ}$ (The tangent at any point of a circle is perpendicular to the radius through the point of contact).
$2$. $OR = OQ$ (Radii of the same circle).
$3$. $OP = OP$ (Common side).
By $R.H.S.$ congruence criterion,$\triangle PRO \cong \triangle PQO$.
Therefore,$\angle RPO = \angle QPO$ (by $CPCT$).
Since $\angle RPO = \angle QPO$,the line $OP$ bisects $\angle RPQ$. Thus,the centre $O$ lies on the angle bisector of the lines $PR$ and $PQ$. Hence proved.