ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid - point of AD.

Solution:

Let EF intersect DB at G.



By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side.

In $\triangle \mathrm{ABD}$,

$E F \| A B$ and $E$ is the mid-point of $A D$.

Therefore, G will be the mid-point of DB.

As EF $\| A B$ and $A B \| C D$

$\therefore \mathrm{EF} \| \mathrm{CD}$ (Two lines parallel to the same line are parallel to each other)

In $\triangle B C D, G F \| C D$ and $G$ is the mid-point of line $B D$. Therefore, by using converse of mid-point theorem, $F$ is the mid-point of $B C$.