Question

In the figure above AB is parallel to $CD,AD=BC$, If $\angle DAB={100}^{\circ }$ find $\angle BCD$

• A. ${80}^{\circ }$
• B. ${60}^{\circ }$
• C. ${70}^{\circ }$
• D. ${50}^{\circ }$

Answer 1

The correct option is A ${80}^{\circ }$

In quadrilateral $ABCD$,

$AB\parallel CD$ and $AD=BC$

$\therefore$ $ABCD$ is an isosceles trapezium.

$\Rightarrow$ $\angle DAB={100}^o$

We know that, in isosceles trapezium sum of opposite angles are supplementary.

$\Rightarrow$ $\angle DAB+\angle BCD={180}^o$

$\Rightarrow$ ${100}^o+\angle BCD={180}^o$

$\Rightarrow$ $\angle BCD={180}^o-{100}^o$

$\therefore$ $\angle BCD={80}^o$