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

(N/A) We have $AB \parallel CD$ and $AD = BC$.
To prove that $\angle A = \angle B$.
Produce $AB$ to $E$ and draw $CE \parallel AD$.
$\therefore AB \parallel DC \Rightarrow AE \parallel DC$ [Given].
Also $AD \parallel CE$ [Construction].
$\therefore AECD$ is a parallelogram.
$\Rightarrow AD = CE$ [Opposite sides of the parallelogram $AECD$].
But $AD = BC$ [Given].
$\therefore BC = CE$.
Now,in $\Delta BCE$,we have $BC = CE$.
$\Rightarrow \angle CBE = \angle CEB$ ......... $(1)$ [$\because$ Angles opposite to equal sides of a triangle are equal].
Also,$\angle ABC + \angle CBE = 180^{\circ}$ [Linear pair] ......... $(2)$.
And $\angle A + \angle CEB = 180^{\circ}$ [$\because$ Adjacent angles of a parallelogram are supplementary] ......... $(3)$.
From $(2)$ and $(3)$,we get $\angle ABC + \angle CBE = \angle A + \angle CEB$.
But $\angle CBE = \angle CEB$ [Using $(1)$].
$\therefore \angle ABC = \angle A$,or $\angle B = \angle A$.