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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