Question
Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.
✓
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. In $\angle\text{RPQ}.$ Construction Join OR, and OQ. In $\triangle\text{POP and }\triangle\text{POO}$ $\angle\text{PRO}\cong\angle\text{PQO}=90^\circ$ [tangent at any point of a circle is perpendicular to the radius through the point of contact] OR = OQ [radii of some circle] Since OP is common $\therefore\ \triangle\text{PRO}\cong\triangle\text{PQO}$ [RHS] Hence, $\angle\text{RPO}=\angle\text{QPO}$ [by CPCT] Thus, O lies on angle bisecter of PR and PQ. Hence proved.
To prove centre of a circle touching two intersecting lines lies on the angle bisector of the lines. In $\angle\text{RPQ}.$ Construction Join OR, and OQ. In $\triangle\text{POP and }\triangle\text{POO}$ $\angle\text{PRO}\cong\angle\text{PQO}=90^\circ$ [tangent at any point of a circle is perpendicular to the radius through the point of contact] OR = OQ [radii of some circle] Since OP is common $\therefore\ \triangle\text{PRO}\cong\triangle\text{PQO}$ [RHS] Hence, $\angle\text{RPO}=\angle\text{QPO}$ [by CPCT] Thus, O lies on angle bisecter of PR and PQ. Hence proved.
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