Question

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

Answer 1

Join $AO$ and extend it to point $B$

In $△APO$

$OP=OA$ (radius)

$\angle OPA=\angle OAP$ (angle opposite to equal sides)

$\angle BOP=\angle OPA+\angle OAP$ (exterior angle property)

$\angle BOP=2\angle OAP\left(1\right)$

In $△AQO$

$OQ=OA$ (radius)

$\angle OQA=\angle OAQ$ (angle opposite to equal sides)

$\angle BOQ=\angle OQA+\angle OAQ$ (exterior angle property)

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

Adding $\left(1\right)$ and $\left(2\right)$

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

$\angle POQ=2\angle PAQ$