Q. 35.0( 28 Votes )
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
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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