In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively

Solution:

$A B C D$ is a parallelogram.

$\therefore \mathrm{AB} \| \mathrm{CD}$

And hence, $A E \|$ FC

Again, $A B=C D$ (Opposite sides of parallelogram $A B C D$ )

$\frac{1}{2} \mathrm{AB}=\frac{1}{2} \mathrm{CD}$

$\mathrm{AE}=\mathrm{FC}(\mathrm{E}$ and $\mathrm{F}$ are mid-points of side $\mathrm{AB}$ and $\mathrm{CD})$

In quadrilateral AECF, one pair of opposite sides (AE and CF) is parallel and equal to each other. Therefore, AECF is a parallelogram.

$\Rightarrow \mathrm{AF} \| \mathrm{EC}$ (Opposite sides of a parallelogram)

In $\triangle \mathrm{DQC}, \mathrm{F}$ is the mid-point of side $\mathrm{DC}$ and $\mathrm{FP} \| \mathrm{CQ}$ (as $\mathrm{AF} \| \mathrm{EC}$ ). Therefore, by using the converse of mid-point theorem, it can be said that $P$ is the mid-point of $D Q$.

$\Rightarrow D P=P Q \ldots(1)$

Similarly, in ΔAPB, E is the mid-point of side AB and EQ || AP (as AF || EC). Therefore, by using the converse of mid-point theorem, it can be said that

$Q$ is the mid-point of $P B$.

$\Rightarrow \mathrm{PQ}=\mathrm{QB} \ldots(2)$

From equations (1) and (2),

$D P=P Q=B Q$

Hence, the line segments AF and EC trisect the diagonal BD.