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Question
ABCD is a cyclic quadrilateral of a circle with centre O such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle. If AD and BC produced meet at P, show that APB = 60°.
Solution
In a circle, ABCD is a cyclic quadrilateral in which AB is the diameter and chord CD is equal to the radius of the circle
To prove – ∠APB = 60°
Construction – Join OC and OD
Proof – Since chord CD = CO = DO ...[Radii of the circle]
∴ ΔDOC is an equilateral triangle
∴ ∠DOC = ∠ODC = ∠DCO = 60°
Let ∠A = x and ∠B = y
Since OA = OB = OC = OD ...[Radii of the same circle]
∴ ∠ODA = ∠OAD = x and ∠OCB = ∠OBC = y
∴ ∠AOD = 180° – 2x and ∠BOC = 180° – 2y
But AOB is a straight line
∴ ∠AOD + ∠BOC + ∠COD = 180°
`=>` 180° – 2x + 180° – 2y + 60° = 180°
`=>` 2x + 2y = 240°
`=>` x + y = 120°
But ∠A + ∠B + ∠P = 180° ...[Angles of a triangle]
`=>` 120° + ∠P = 180°
`=>` ∠P = 180° – 120°
`=>` ∠P = 60°
Hence ∠APB = 60°
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