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# 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.