Question

In the given figure, $ABCD$ is a cyclic quadrilateral. $AF$ is drawn parallel to $CB$ and $DA$ is produced to point $E$. If $\angle ADC={92}^{\circ }$, $\angle FAE={20}^{\circ }$; determine $\angle BCD$.

Answer 1

Given,
In cyclic quad. $ABCD$
$AF||CB$ and $DA$ is produced to $E$ such that $\angle ADC={92}^{\circ }$ and $\angle FAE={20}^{\circ }$
So,
$\angle B+\angle D={180}^{\circ }$
$\angle B+{92}^{\circ }={180}^{\circ }$
$\angle B={88}^{\circ }$
As $AF||CB$,

$\angle FAB=\angle B={88}^{\circ }$ {Altenate angles}
But, $\angle FAE={20}^{\circ }$ [Given]
$\angle A+\angle BAF+\angle FAE={180}^o$ {Angle on a straight line}
$\angle A+{88}^{\circ }+{20}^{\circ }={180}^{\circ }$............(1)
Also,

$\angle A+\angle C={180}^o$ ............ (2) {Cyclic quadrilateral}
Subtracting (1) from (2) we get,

$\angle C={108}^o$

$\therefore \angle BCD={108}^{\circ }$