Question

In the following figure,
If $\angle BAD={96}^o$, find $\angle BCD$ and $\angle BFE$

Answer 1

$ABCD$ is a cyclic quadrilateral.

$\therefore \angle BAD+\angle BCD=180°(\because \text{In a cyclic quadrilateral, pair of opposite angles are supplementary.})$

$\Rightarrow \angle BCD=180°-96°=84°$

$\therefore \angle BCE=180°-\angle BCE=180°-84°=96°$

Similarly, $BCEF$ is a quadrilateral.

$\angle BCE+\angle BFE=180°$

$\Rightarrow \angle BFE=180°-96°=84°$