$ABCD$ is a trapezium in which $AB \parallel CD$ and $AD = BC$ (see Fig). Show that $\angle C = \angle D$.

(N/A) To prove that $\angle C = \angle D$.
Construction: Extend $AB$ to $E$ and draw a line through $C$ parallel to $AD$ intersecting $AB$ produced at $E$.
Since $AD \parallel CE$ and $AE \parallel DC$,$AECD$ is a parallelogram.
Therefore,$AD = CE$ (opposite sides of a parallelogram).
Given that $AD = BC$,it follows that $BC = CE$.
In $\triangle BCE$,since $BC = CE$,the angles opposite to these sides are equal,so $\angle CBE = \angle CEB$.
Also,$\angle ABC + \angle CBE = 180^{\circ}$ (linear pair).
Since $AECD$ is a parallelogram,$\angle A = \angle ADC$ and $\angle D + \angle A = 180^{\circ}$ (consecutive interior angles).
Also,$\angle CEB = \angle A$ (corresponding angles as $AD \parallel CE$).
Since $\angle D + \angle A = 180^{\circ}$ and $\angle C + \angle B = 180^{\circ}$,and using the properties of the parallelogram and isosceles triangle,we conclude $\angle C = \angle D$.