# Prove that the intercept of a tangent between two parallel tangents to a circle subtendsa right angle at the centre.

To do:

We have to prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.

Solution:

Let $PQ$ and $RS$ be the parallel tangents of a circle.

$RMP$ is the intercept of the tangent between $PQ$ and $RS$.
Join $RO$ and $PQ$, where $O$ is the centre of the circle.

$RL$ and $RM$ are the tangents and $\mathrm{RO}$ is joined.

This implies,

$\angle LRO=\angle MRO$.......(i)

Similarly,

$PM$ and $PN$ are the tangents and $\mathrm{PO}$ is joined.

$\angle NPO=\angle MPO$.......(ii)

Adding equations (i) and (ii), we get,

$\angle LRO+\angle NPO=\angle MRO+\angle MPO$

$\angle \mathrm{LRM}+\angle \mathrm{MPN}=180^{\circ}$   (co-interior angles)

$\Rightarrow \angle LRO+\angle MRO+\angle MPO+\angle NPO=180^{\circ}$

$\Rightarrow \angle MRO+\angle MRO+\angle MPO+\angle MPO=180^{\circ}$

$\Rightarrow 2(\angle MRO+\angle MPO)=180^{\circ}$

$\Rightarrow \angle MRO+\angle MPO=\frac{180^{\circ}}{2}=90^{\circ}$

In $\Delta \mathrm{POR}$,

$\angle MRO+\angle MPO+\angle POR=180^{\circ}$

$\Rightarrow 90^{\circ}+\angle POR=180^{\circ}$

$\Rightarrow \angle POR=180^{\circ}-90^{\circ}$

$\angle POR=90^{\circ}$

Hence proved.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

38 Views