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Sum of Opposite Angles of a Cyclic Quadrilateral

In the given ...

Question

In the given figure, ABCD is a cyclic quadrilateral with O as the centre of the circle. If ∠BOD = 136°, find ∠BCD.

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Solution

$We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point in the remaining part of the circle . So, ∠ B A D = \frac{1}{2} ∠ B O D \Rightarrow ∠ B A D = \frac{1}{2} \times 136 ° = 68 °$
$∠ B C D + ∠ B A D = 180 ° (A B C D is a cyclic quadrilateral) \Rightarrow ∠ B C D = 180 ° - ∠ B A D = 180 ° - 68 ° = 112 °$

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