Question

Prove that the line joining the mid-points of the diagonals of a trapezium is parallel tot he parallel sides of the trapezium.

Answer 1

Given: A trapezium $ABCD$ in which $E$ and $F$ are the mid-points of sides $AD$ and $BC$ respectively.

To prove: $EF||AB||DC$

Construction: Join $DE$ and produce it to intersect $AB$ produced at $G$.

Proof: In $\Delta DCF$ and $\Delta GBF$, we have

$\angle 1=\angle 2$ [Alternate interior angles because $DC|AG$]

$\angle 3=\angle 4$ [Vertically opposite angles]

$CF=BF$ [$\because F$ is the mid-point of $BC$]

So, By AAS criterion of congruence, we have

$\Delta DCF\cong \Delta GBF$
$\therefore DF=GF\left[CPCT\right]$

In $\Delta DAG,EF$ joins mid-points of sides $DA$ and $DG$ respectively

$\therefore EF||AG$ [Mid-point theorem]

$\Rightarrow EF||AB$

But, $AB||DC$ [Given]

$\therefore EF||BC||DC$