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 line s lies on the angle bisector of the lines.
Construction: Join OR, and OQ.
To prove: Centre of a circle touching two intersecting line s lies on the angle bisector of the lines.
Construction: Join OR, and OQ.
In $\triangle\text{POR}\ \text{and}\ \triangle\text{PQO}$
$\angle\text{PRO}=\angle\text{PQO}=90^{\circ}$ [tangent at any point of a circle is perpendicular to the radius through the point of contact]
OR = OQ [radii of same circle]
Since, OP is common.
$\triangle\text{PRO}\cong\triangle\text{PQO}$ [SAS]
Hence, $\angle\text{RPO}=\angle\text{QPO}$ [by CPCT]
Thus, O lies on angle bisector of PR and PQ.
Hence proved.
$\angle\text{PRO}=\angle\text{PQO}=90^{\circ}$ [tangent at any point of a circle is perpendicular to the radius through the point of contact]
OR = OQ [radii of same circle]
Since, OP is common.
$\triangle\text{PRO}\cong\triangle\text{PQO}$ [SAS]
Hence, $\angle\text{RPO}=\angle\text{QPO}$ [by CPCT]
Thus, O lies on angle bisector of PR and PQ.
Hence proved.
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