Question

In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). 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)

`1/2  AB` = `1/2 CD`

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 ΔDQC, 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 ΔAPB, 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.