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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