Can anyone give a proof for ∠AOB =2∠APB of a circle

Question

We know that angle subtended by an arch is twice at the centre than anywhere else in the circle. So if the angle at centre is AOB(say) so can anyone prove that angle on the circle subtended (say APB) by the same arch is half of it. Couldn't add diagram cos I'm new.

Answer 1

Note that $\angle AOC = 2 \angle APO$ and $\angle BOC = 2 \angle BPO$ because the triangles APO and BPO are isosceles. Thus, $\angle AOB = 2 \angle APB$.

Answer 2

Connect $O$ to $A,B,P$.

Since $AOP.BOP$ and $OAB$ are isosceles, you ahve $$\angle OPA =\angle OAP \\ \angle OPB =\angle OBP$$

Now, $$\angle POA + \angle POB+\angle AOB = 360^\circ = \angle OPA+\angle OAP +\angle POA + \angle OPB+\angle POB+\angle PBO$$

Canceling you get $$ \angle AOB = \angle OPA+\angle OAP + \angle OPB+\angle PBO =2 \angle OPA+ 2 \angle OPB$$