Question -
Answer -
Given : Two tangents PQ and PR are drawn from an external point P to a circle with centre O.
To prove : Centre of a circle touching two intersecting lines lies on the angle bisector of the lines.
Construction : Join OR, and OQ.
In ∆POR and ∆POQ
∠PRO = ∠PQO = 90°
[tangent at any point of a circle is perpendicular to the radius through the point of contact]
OR = OQ [radii of same circle]
Since, OP is common.
∆PRO = ∆PQO [RHS]
Hence, ∠RPO = ∠QPO [by CPCT]
Thus, O lies on angle bisector of PR and PQ.
Hence proved.