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]
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.