Question

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.

Answer 1

ABCD is a parallelogram.

∠AB || CD

And hence, AE || FC

Again, AB = CD (Opposite sides of parallelogram ABCD)

AB= $\frac{1}{2}$ CD= $\frac{1}{2}$

AE = FC (E and F are mid-points of side AB and CD)

In quadrilateral AECF, one pair of opposite sides (AE and CF) is parallel and equal to each other. Therefore, AECF is a parallelogram. ∠AF || EC (Opposite sides of a parallelogram)

In ∆DQC, F is the mid-point of side DC and FP || CQ (as AF || EC). Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point of

DQ.

∠DP = PQ ... (1)

Similarly, in ∆APB, E is the mid-point of side AB and EQ || AP (as AF || EC).

Therefore, by using the converse of mid-point theorem, it can be said that Q is the mid-point of PB.

∠PQ = QB ... (2)

From equations (1) and (2), DP =

PQ = BQ

Hence, the line segments AF and EC trisect the diagonal BD.