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

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 .

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