Let PR and PQ are two intersecting lines [intersects on point P] touching the circle with center O and we joined OR, OQ and OP.

To Prove : Center O lies on the angle bisector of PR and PQ i.e. OP is the angle bisector of ∠RPQ

Proof :

Clearly, PQ and PQ are tangents to the circle with a common external point P.

In △POR and △POQ

OR = OQ [radii of same circle]

OP =OP [Common]

PR = PQ [Tangents drawn from an external point to a circle are equal ]

△POR ≅ △POQ [ By Side Side Side criterion ]

∠RPO = ∠OPQ [ Corresponding parts of congruent triangles are equal ]

This implies that OP is the angle bisector of ∠RPQ .

Hence Proved .