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

Given:

The centre of a circle touching two intersecting lines.

To do:
We have to prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

 Solution:

Let two tangents $PQ$ and $PR$ are drawn from an external point $P$ to a circle with centre $O$.

Join $OR$ and $OQ$.

In $\triangle POR$ and $\triangle POQ$,

$\angle PRO = \angle PQO = 90^o$      (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

$OR = OQ$    (Radii of the circle)

$OP = OP$     (Common side)

Therefore, by RHS congruency,

$\triangle PRO\ \cong\ \triangle PQO$

This implies,

$\angle RPO = \angle QPO$  ( CPCT)

Hence, $O$ lies on the angle bisector of $PR$ and $PQ$.

Hence proved.

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Updated on: 10-Oct-2022

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