Q. 84.0( 4 Votes )

In the picture, A

Answer :

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.


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