**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.**