Question

Prove that the angle subtended by an arc at the centre is double the angel subtended by it at any point on the remaining part of the circle.

Answer 1

Case $I$

In $△APO,OP=OA$(radius)

$\angle OPA=\angle OAP$(angles opposite to equal sides are equal) ....$\left(1\right)$

In $△AQO,OQ=OA$(radius)

$\angle OQA=\angle OAQ$(angles opposite to equal sides are equal) ....$\left(2\right)$

Also, by exterior angle property,Exterior angle is sum of interior opposite angles

$\angle BOP=\angle OPA+\angle OAP$

$\angle BOP=\angle OAP+\angle OAP$

$\angle BOP=2\angle OAP$ ....$\left(3\right)$

Also, by exterior angle property,Exterior angle is sum of interior opposite angles

$\angle BOQ=\angle OQA+\angle OAQ$

$\angle BOQ=\angle OAQ+\angle OAQ$

$\angle BOQ=2\angle OAQ$ ....$\left(4\right)$

Adding $\left(3\right)$ and $\left(4\right)$ we get

$\angle BOP+\angle BOQ=2(\angle OAP+\angle OAQ)$

Hence $\angle POQ=2\angle PAQ$

Hence proved.

Case $II$

In $△APO,OP=OA$(radius)

$\Rightarrow \angle OPA=\angle OAP$(Angles opposite to equal angles are equal) .....$\left(1\right)$

In $△AQO,OQ=OA$(radius)

$\Rightarrow \angle OQA=\angle OAQ$(Angles opposite to equal angles are equal) .....$\left(2\right)$

Also, by exterior angle property,Exterior angle is sum of interior opposite angles

$\angle BOP=\angle OPA+\angle OAP$

$\angle BOP=\angle OAP+\angle OAP$

$\angle BOP=2\angle OAP$ ....$\left(3\right)$

Also, by exterior angle property,Exterior angle is sum of interior opposite angles

$\angle BOQ=\angle OQA+\angle OAQ$

$\angle BOQ=\angle OAQ+\angle OAQ$

$\angle BOQ=2\angle OAQ$ ....$\left(4\right)$

Adding $\left(3\right)$ and $\left(4\right)$ we get

$\angle BOP+\angle BOQ=2(\angle OAP+\angle OAQ)$

Hence reflex angle $\angle POQ=2\angle PAQ$

${360}^{\circ }-\angle POQ=2\angle PAQ$

Hence proved.