Question

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

Answer 1

$ABCD$ is a trapezium in which $AB\parallel DC$ and $E,F$ are mid points of $AD,BC$ respectively.

Join $CE$ and produce it to meet $BA$ produced at $G$.

In $△EDC$ and $△EAG$,

$ED=EA$ ( $E$ is mid point of $AD$)

$\angle CED=\angle GEC$ ( Vertically opposite angles)

$\angle ECD=\angle EGA$ ( alternate angles) ( $DC\parallel AB$, $DC\parallel GB$ and $CG$ is the transversal)

$\therefore △EDC\cong △EAG$

$CD=GA$ and $EC=EG$

In $△CGB$,

$E$ is mid point of $CG$ ( $EC=EG$ proved)

$F$ is a mid point of $BC$ (given)

$\therefore$ By mid point theorem $EF\parallel AB$ and $EF=\frac{1}{2}GB$