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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.

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Given : A parallelogram ABCD, in which E and F are mid-points of sides AB and DC respectively. 

To Prove : DP = PQ = QB 

Proof : Since E and F are mid-points of AB and DC respectively. 

⇒ AE = 1/ 2 AB and CF = 1 /2 DC …(i) 

But, AB = DC and AB || DC …(ii) [Opposite sides of a parallelogram] 

∴ AE = CF and AE || CF. 

⇒ AECF is a parallelogram. [One pair of opposite sides is parallel and equal] 

In ∆BAP, 

E is the mid-point of AB 

EQ || AP 

⇒ Q is mid-point of PB [Converse of mid-point theorem] 

⇒ PQ = QB …(iii) 

Similarly, in ∆DQC, 

P is the mid-point of DQ 

DP = PQ …(iv) 

From (iii) and (iv), we have 

DP = PQ = QB 

or line segments AF and EC trisect the diagonal BD. Proved.

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