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. Prove that the angle which the small arc AB makes at O is the sum of the angles it makes at P and Q.

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.