$ABCD$ is a quadrilateral in which $AB \parallel DC$ and $AD = BC$. Prove that $\angle A = \angle B$ and $\angle C = \angle D$.

(N/A) Given: $A$ quadrilateral $ABCD$ in which $AB \parallel CD$ and $AD = BC$.
To prove: $\angle A = \angle B$ and $\angle C = \angle D$.
Construction: Draw $DP \perp AB$ and $CQ \perp AB$.
Proof: In $\triangle APD$ and $\triangle BQC$,we have:
$\angle 1 = \angle 2 = 90^{\circ}$ (By construction)
$AD = BC$ (Given)
$DP = CQ$ (Distance between parallel lines is constant)
By $RHS$ criterion of congruence,we have:
$\triangle APD \cong \triangle BQC$ (By $CPCT$)
$\therefore \angle A = \angle B$
Now,since $DC \parallel AB$:
$\angle A + \angle D = 180^{\circ}$ (Sum of consecutive interior angles is $180^{\circ}$)
$\angle B + \angle C = 180^{\circ}$ (Sum of consecutive interior angles is $180^{\circ}$)
Since $\angle A = \angle B$,we have:
$180^{\circ} - \angle A = 180^{\circ} - \angle B$
$\Rightarrow \angle D = \angle C$
Hence,proved.