In the picture, A

Given: In the picture, A, B, C, D are points on a circle centred at O. The line AC and BD are extended to meet at P. The line AD and BC intersect at Q

To Prove: The angle which the small arc AB makes at O is the sum of the angles it makes at P and Q i.e.

∠AOB = ∠APB + ∠AQB

We know that,

The angle made by an arc of circle on the alternate arc is half the angle made at center.

Therefore,

…[1] and

…[2]

Also, By linear pair

∠ACB + ∠PCB = 180°

∠ACB = 180° – ∠PCB …[3]

And

∠ADB + ∠ADP = 180°

∠ADB = 180° – ∠ADP …[4]

Also, By angle sum property of quadrilateral of CQDP

∠CQD + ∠QCP + ∠CPD + ∠PDQ = 360°

⇒ ∠AQB + ∠PCB + ∠APB + ∠ADP = 360°

⇒ ∠PCB + ∠ADP = 360°– (∠AQB + ∠APB) …[5]

[Here, ∠CQD = ∠AQB, vertically opposite angles]

Adding [1] and [2]

⇒ ∠ACB + ∠ADB = ∠AOB

⇒ 180° – ∠PCB + 180° – ∠ADP = ∠AOB [Using [3] and [4]]

⇒ 360° – (∠PCB + ∠ADP) = ∠AOB

⇒ 360° – (360° – (∠AQB + ∠APB)) = ∠AOB

⇒ ∠AOB = ∠AQB + ∠APB

Hence Proved.