The line segment joining the mid-points $M$ and $N$ of the parallel sides $AB$ and $DC$ respectively of a trapezium $ABCD$ is perpendicular to both the sides $AB$ and $DC$. Prove that $AD = BC$.

(N/A) Join $MD$ and $MC$.
In $\Delta DMN$ and $\Delta CMN$:
$DN = CN$ [Since $N$ is the mid-point of $DC$]
$\angle DNM = \angle CNM = 90^{\circ}$ [Given]
$MN = MN$ [Common side]
Therefore,$\Delta DMN \cong \Delta CMN$ [By $SAS$ congruence rule]
This implies $DM = CM$ and $\angle NMD = \angle NMC$ ... $(1)$ [$CPCT$]
Now,consider $\Delta AMD$ and $\Delta BMC$:
$AM = BM$ [Since $M$ is the mid-point of $AB$]
$DM = CM$ [From $(1)$]
$\angle AMD = \angle AMN - \angle NMD$
$\angle BMC = \angle BMN - \angle NMC$
Since $\angle AMN = \angle BMN = 90^{\circ}$ and $\angle NMD = \angle NMC$,we have $\angle AMD = \angle BMC$ ... $(2)$
Therefore,$\Delta AMD \cong \Delta BMC$ [By $SAS$ congruence rule]
Thus,$AD = BC$ [$CPCT$].