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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.
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