Question

In the given figure, ABCD is a cyclic quadrliateral in which AE is drawn parallel to CD, and BA is produced to F. If $\angle ABC={92}^{\circ }and\angle FAE={20}^{\circ },\text{find}\angle BCD.$

Answer 1

ANSWER:
Given: ABCD is a cyclic quadrilateral.
Then ∠ ABC + ∠ ADC = 180 °⇒ 92 ° + ∠ ADC = 180 °
⇒ ∠ ADC = (180 ° – 92 ° ) = 88 °
Again, AE parallel to CD.
Thus, ∠ EAD = ∠ ADC = 88 ° (Alternate angles)
We know that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
∴ ∠ BCD = ∠ DAF
⇒ ∠ BCD = ∠ EAD + ∠ EAF
= 88 ° + 20 ° = 108 °
Hence, ∠ BCD = 108 °