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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. AF is drawn parallel to CB and DA is produced to point E. If $∠ A D C = 92^{\circ}, ∠ F A E = 20^{\circ}; d e t e r m i n e ∠ B C D .$ Give reason in support of your answer.

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Solution

Given: $ABCD$ is a cyclic quadrilateral

$AF \parallel CB. \angle ADC = 92^o$ and $\angle FAE = 20^o$

$\angle CDA + \angle CBA = 180^o$

$\angle CBA = 180-92 = 88^o$

Since $AF \parallel CB$

$\angle CBA = \angle BAF$ (alternate angles)

$\angle BAE = 88 + 20 = 108^o$

Therefore $a n g l e D A B = 180^{0} - 108^{0} = 72^{0}$

Therefore $\angle BCD = 180-72 = 108^o$

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