Question

$ABCD$ is a trapezium in which $AB\parallel CD$ and $\angle A=\angle B={45}^0.$ Find $\angle C$ and $\angle D$ of this trapezium.

Answer 1

$ABCD$ is a trapezium in which $AB||DC$

Refer image,

Now, $AB||DC$ and $AD$ is transversal.

$\therefore \angle A+\angle D={180}^0$

($\because$ Sum of interior on the side of the transversal is ${180}^0$)

$\Rightarrow {45}^0+\angle D={180}^0$

$\Rightarrow \angle D={180}^0-{45}^0={135}^0$

Similarly, $\angle B+\angle C={180}^0$

$\Rightarrow {45}^0+\angle D={180}^0$

$\Rightarrow \angle C={180}^0-{45}^0={135}^0$

Hence, $\angle A=\angle B={45}^0$ and $\angle C=\angle D={135}^0$