Question

In a parallelogram $ABCD$, the side $DC$ is produced to $E$ and $\angle BCE={105}^{\circ }$. Calculate $\angle A,\angle B,\angle C$, and $\angle D$.

Answer 1

We can see that,
$\angle BCD$ + $\angle BCE$ = $180°$ ($\because$ Linear pair)
so, $\angle BCD$ = ${180}^{\circ }$ - ${105}^{\circ }$ = ${75}^{\circ }$
In $ABCD$, $\angle A$ =$\angle C$ = ${75}^{\circ }$ (Opposite angles of parallelogram)
$\angle ABC$ = $\angle BCE$ = ${105}^{\circ }$ (Alternate angles, $AB||DE$)
$\angle D$ = $\angle B$ = ${105}^{\circ }$ (Opposite angles of parallelogram are equal )