Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium.
Given Let ABCD be a trapezium in which AB|| DC and let M and N be the mid-points of the diagonals AC and BD, respectively.
To prove $M N\|A B\| C D$
Construction Join $C N$ and produce it to meet $A B$ at $E$.
In $\triangle C D N$ and $\triangle E B N$, we have
$D N=B N$ [since, $N$ is the mid-point of $B D$ ]
and $\angle D C N=\angle B E N$ [alternate interior angles]
$\begin{array}{ll}\text { and } & \angle C D N=\angle E B N \quad \text { [alternate interior angles] }\end{array}$
$\therefore$ $D C=E B$ and $C N=N E$ [by CPCT rule]
Thus, in $\triangle C A E$, the points $M$ and $N$ are the mid-points of $A C$ and $C E$, respectively.
$\therefore \quad M N \| A E \quad$ [by mid-point theorem]
$\Rightarrow$ $M N\|A B\| C D$ Hence proved.
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