Question

$E$ and $F$ are respectively the midpoint of non-parallel sides $AD$ and $BC$ of a trapezium $ABCD$, prove that $EF$ $||ABandEF=\frac{1}{2}(AB+CD)$

Answer 1

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

To prove: $EF||AB$ and $EF=\frac{1}{2}(AB+CD)$

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

Proof: In $\Delta CFD$ and $\Delta BFG$, we have

$DC||AB$

$\therefore \angle C=\angle 3$ [ALternate interior angles]

$CF=BF$

$\angle 1=\angle 2$ [Vertically opposite angles]

So, by ASA criterion of congruence, we have

$\Delta CFD\cong \Delta BFG$

$\therefore CD=BG\left(CPCT\right)$

In $\Delta DAG,EF$ joins mid-points of sides $AD$ and $GD$ respectively

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

$\Rightarrow EF||AB$

So, $EF=\frac{1}{2}AG$[ Mid-point theorem]

$EF=\frac{1}{2}(AB+AG)$

$EF=\frac{1}{2}(AB+CD)[\because CD=BG]$

Hence, proved.