In quadrilateral ABCD,
AB = CD [∵ Given]
(1/2)AB = (1/2)CD
⇒ AE = CF [∵ E and F are the mid-points of AB and CD respectively]
In Quadrilateral AECF,
AE = CF [∵ Proved above]
AE ∥ CF [∵ Opposite sides of a parallelogram]
Hence, AECF is a parallelogram.
In ΔDCQ,
F is mid-point of DC [∵ Given]
and FP ∥ CQ [∵ AECF is a Parallelogram]
Hence, P is mid-point of DQ [ Converse of mid-point theorem]
Hence, DP = PQ …(1)
Similarly,
In ΔABP,
E is mid-point of AB [∵ Given]
and EQ ∥ AP [∵ AECF is a parallelogram]
Hence, Q is mid-point of PB [∵ Converse of mid-Point Theorem]
Hence, PQ = QB …(2)
From (1) and (2), we have
DP = PQ = QB
Hence, line segment AF and EC trisect BD.