the sum of either pair of opposite angles of a cyclic quadrilateral is 180°
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Let ABCD be a cyclic quadrilateral with center of the circle at O.
Join BO and DO.
Arc DAB makes ∠2x at center O and hence ∠x (half of angle at O) at C.
Arc DCB makes ∠2y at O and hence ∠y (half of angle at O) at A.
We know , ∠2x + 2y = 360° => ∠x + ∠y = 180°
Hence, ∠A + ∠C = x+y = 180°
So ∠B + ∠D = 360 - ∠A + ∠C = 180°
Join BO and DO.
Arc DAB makes ∠2x at center O and hence ∠x (half of angle at O) at C.
Arc DCB makes ∠2y at O and hence ∠y (half of angle at O) at A.
We know , ∠2x + 2y = 360° => ∠x + ∠y = 180°
Hence, ∠A + ∠C = x+y = 180°
So ∠B + ∠D = 360 - ∠A + ∠C = 180°
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