Given: ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively.
To prove: AF and EC trisect the diagonal BD
ABCD is a parallelogram
AB || CD
AE || FC
AB = CD (Opposite sides of parallelogram ABCD)
⇒½ AB = ½ CD
⇒ AE = FC (E and F are midpoints of side AB and CD)
AECF is a parallelogram (AE and CF are parallel and equal to each other)
AF || EC (Opposite sides of a parallelogram)
From ΔDQC, F is mid point of side DC and FP || CQ (as AF || EC).
P is the mid-point of DQ (Converse of mid-point theorem)
⇒ DP = PQ — (i)
E is midpoint of side AB and EQ || AP (as AF || EC).
Q is the mid-point of PB (Converse of mid-point theorem)
⇒ PQ = QB — (ii)
From equations (i) and (ii),
DP = PQ = BQ
Therefore, the line segments AF and EC trisect the diagonal BD.