Question

In the given figure, ABCD is a cyclic quadrilateral, AE is drawnn parallel to CD and BA is produced. If $\angle ABC={92}^{\circ }$ and $\angle FAE={20}^{\circ }$, then $\angle BCD$ is equal to

Answer 1

We know that $ABCD$ is a cyclic quadrilateral

So we get

$\angle ABC+\angle ADC={180}^o$

By substituting the values

${92}^o+\angle ADC={180}^o$

On further calculation

$\angle ADC={180}^o-{92}^o$

By subtraction

$\angle ADC={88}^o$

We know that $AE\parallel CD$

From the figure we know that

$\angle EAD=\angle ADC={88}^o$

We know that the exterior angle of a cyclic quadrilateral = interior opposite angle

So we get

$\angle BCD=\angle DAF$

We know that

$\angle BCD=\angle EAD+\angle EAF$

It is given that $\angle FAE={20}^o$

By substituting the values

$\angle BCD={88}^o+{20}^o$

By addition

$\angle BCD={108}^o$

Therefore, $\angle BCD={108}^o$.