Question
In the given figure, $ABCD$ is a cyclic quadrilateral in which $AE$ is drawn parallel to $CD$, and $BA$ is produced. If $\angle\text{ABC}=92^\circ$ and $\angle\text{FAE}=20^\circ,$ find $\angle\text{BCD}.$
✓
Answer
Given: $ABCD$ is a cyclic quadrilateral.
Then $\angle\text{ABC}+\angle\text{ADC}=180^\circ$
$\Rightarrow\ 92^\circ+\angle\text{ADC}=180^\circ$
$\Rightarrow\ \angle\text{ADC}=(180^\circ-92^\circ)=88^\circ$ Again, $AE$ parallel to $CD$.
Thus, $\angle\text{EAD}=\angle\text{ADC}=88^\circ$ [Alternate angles]
We know that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
$\therefore\ \angle\text{BCD}=\angle\text{DAF}$
$\Rightarrow\ \angle\text{BCD}=\angle\text{EAD}+\angle\text{EAF}$
$=88^\circ+20^\circ=108^\circ$
Hence, $\angle\text{BCD}=108^\circ$
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