Question

$E$ and $F$ are respectively the midpoints of non-parallel sides $AD,BC$ of trapezium $ABCD.$ Prove that $EF\parallel AB.$

Answer 1

$ABCD$ is a trapezium in which $AB||DC$ and $E,F$ are mid point of $AD,BC$ respectively. Join $CE$ and produce it to meet $BA$ produced at $G.$

In $△EDC$ and $△EAG,$

$ED=EA$ [$E$ is the mid point of $AD.$]

$\angle CED=\angle GEC$ [Vertically opposite angles]

$\angle ECD=\angle EGA$ [Alternating angles]

So, by $ASA$ criteria of similarity,

$△EDC\cong △EAG$

$\Rightarrow CD=GA$ and $EC=EG$ which implies that $E$ is the midpoint of $CG$ as well.

In $△CGB,$

$E$ is a mid point $CG$ and $F$ is a midpoint of $BC.$

$\Rightarrow \frac{CE}{EG}=\frac{CF}{FB}=1$

So, by the converse of basic proportionality theorem $EF\parallel AB.$