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Show that the line segments AF and EC trisect the diagonal BD.

Answer :

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.

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