In a parallelogram $ABCD,E$ and $F$ are the mid-points of sides $AB$ and $CD$ respectively (see Fig). Show that the line segments $AF$ and $EC$ trisect the diagonal $BD$.

$ABCD$ is parallelogram

Using properties of a parallelogram $AB\parallel DC$ and $AB=DC$

$E$ is the mid-point of $AB$
$AE=\frac{1}{2}AB$    ... (1)

$F$ is the mid-point of $CD$
Therefore, $CF=\frac{1}{2}CD$
$CF=\frac{1}{2}AB$ (Since $CD=AB$)    ... (2)

From (1) and (2), we get

$AE=CF$
Also,

$AE\parallel CF$ (Since $AB\parallel DC$)

Thus, a pair of opposite sides of a quadrilateral  $AECF$ are parallel and equal.

Quadrilateral $AECF$ is a parallelogram.

$EC\parallel AF$
$EQ\parallel AP$ and $QC\parallel PF$

In $△BPA$,  $E$ is the mid-point  of $BA$         

$EQ\parallel AP$ $\mid$ 
Using mid point theorem $BQ=PQ$          ... (3)

Similarly, by taking  $△CQD$,  we can prove that

$DP=QP$    ... (4)
From (3) and (4), we get

$BQ=QP=PD$

Therefore,

$AF$ and $CE$ trisect the diagonal $BD$.