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

Given: ABCD is a parallelogram and E, F are the mid-points of sides AB and CD, respectively.

To Prove: Line segments AF and EC trisect the diagonal BD.

Proof:
\(\because \) ABCD is a parallelogram,
\(\therefore \) AB || DC and AB = DC
(since, Opposite sides of a parallelogram)
Thus, AE || FC and (\(\frac{1}{2} \)) AB = (\(\frac{1}{2} \)) CD
\(\Rightarrow \) AE = FC
\(\therefore \) AECF is a parallelogram.
Thus, AF || EC
And hence, EQ || AP and FP || CQ.

In \(\triangle{BAP}\), E is the mid-point of AB and EQ || AP,
So, By converse of mid-point theorem,
Q is the mid-point of BP.
\(\Rightarrow \) BQ = PQ ...(i)

Similarly, in \(\triangle{DQC}\), F is the mid-point of DC and FP || CQ,
So, P is the mid-point of DQ.
\(\Rightarrow \) DP = PQ ...(ii)
From Equations (i) and (ii) , we get,
BQ = PQ = DP

Hence, it is proved that CE and AF trisect the diagonal BD.