Prove that the line segment joining the midpoints of the diagonals of a trapezium is parallel to the parallel sides of the trapezium and is equal to half the difference of the lengths of the parallel sides.

(N/A) Let $ABCD$ be a trapezium with $AB \parallel DC$. Let $P$ and $Q$ be the midpoints of the diagonals $AC$ and $BD$ respectively.
$1$. Join $AQ$ and extend it to meet $DC$ at point $E$.
$2$. In $\triangle ABQ$ and $\triangle EDQ$:
- $\angle ABQ = \angle EDQ$ (Alternate interior angles as $AB \parallel DE$)
- $BQ = DQ$ ($Q$ is the midpoint of $BD$)
- $\angle AQB = \angle EQD$ (Vertically opposite angles)
By $ASA$ congruence criterion,$\triangle ABQ \cong \triangle EDQ$.
$3$. Therefore,$AQ = EQ$ and $AB = DE$ (by $CPCT$).
$4$. In $\triangle ACE$,$P$ is the midpoint of $AC$ and $Q$ is the midpoint of $AE$ (since $AQ = EQ$).
$5$. By the Midpoint Theorem,$PQ \parallel CE$ and $PQ = \frac{1}{2} CE$.
$6$. Since $CE = DC - DE = DC - AB$,we have $PQ = \frac{1}{2} (DC - AB)$.
$7$. Thus,$PQ$ is parallel to the parallel sides $AB$ and $DC$ and its length is half the difference of the lengths of the parallel sides.