The angle subtended by an arc …

Solution: Given : Arc AB of a circle with center O, Subtending $\angle AOB$ at the centre and $\angle APB$ at a point P on the remaining part of circle.

To Prove : $\angle AOB = 2\angle APB$

Proof :

We can have three cases;

i) AB is a minor arc

ii) AB is a diameter 

iii) AB is a major arc

we are taking first case only at this time.

We know that, an exterior angle of a triangle is equal to the sum of the interior opposite angles.

$In ∆OPB, \angle QOB = \angle OPB + \angle OBP ..... \left(1\right)$

OB = OP (Radius of the circle)

=> $\angle OPB = \angle OBP$ (In a triangle, equal sides have equal angle opposite to them)

=> $\angle QOB = \angle OPB + \angle OPB$

=> $\angle QOB = 2\angle OPB ...... \left(2\right)$

$in ∆OPA, \angle QOA = \angle OPA + \angle OAP ........\left(3\right)$

OA = OP (Radius of the circle)

=> $\angle OPA=\angle OAP$   (In a triangle, equal sides have equal angle opposite to them)

=> $\angle QOA = \angle OPA + \angle OPA$

=> $\angle QOA = 2\angle OPA ......\left(4\right)$

Adding (2) and (4), We get 

$\angle QOB + \angle QOA = 2\angle OPB + 2\angle OPA$

=> $\angle AOB=2\angle APB$

Hence Proved