$ABCD$ is a trapezium in which $AB \parallel DC$ and $\angle A = \angle B = 45^{\circ}$. Find angles $C$ and $D$ of the trapezium.

(N/A) $ABCD$ is a trapezium in which $AB \parallel DC$.
Since $AB \parallel DC$ and $AD$ is a transversal,the sum of interior angles on the same side of the transversal is $180^{\circ}$.
Therefore,$\angle A + \angle D = 180^{\circ}$.
Substituting the value of $\angle A = 45^{\circ}$:
$45^{\circ} + \angle D = 180^{\circ}$
$\angle D = 180^{\circ} - 45^{\circ} = 135^{\circ}$.
Similarly,since $AB \parallel DC$ and $BC$ is a transversal,the sum of interior angles on the same side of the transversal is $180^{\circ}$.
Therefore,$\angle B + \angle C = 180^{\circ}$.
Substituting the value of $\angle B = 45^{\circ}$:
$45^{\circ} + \angle C = 180^{\circ}$
$\angle C = 180^{\circ} - 45^{\circ} = 135^{\circ}$.
Thus,the angles are $\angle C = 135^{\circ}$ and $\angle D = 135^{\circ}$.