Question

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

Answer 1

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

In $△EDC$ and $△EAG,$

$\Rightarrow$ $\angle CED=\angle GEA$ [ Vertically opposite angles ]

$\Rightarrow$ $\angle ECD=\angle EGA$ [ Alternate angles ]

$\Rightarrow$ $ED=EA$ [ Since, $E$ is the midpoint of $AD$ ]

$\Rightarrow$ $△EDC\cong △EAG$ [ By AAS congruence theorem ]

$\Rightarrow$ $CD=GA$ and $EC=EG$ [ By CPCT ]

In $△CGB,$

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

By mid-point theorem,

$\therefore$ $EF\parallel AB$

Hence Proved.