In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.

To Prove: AF and EC trisect the diagonal BD.

Given: E and F are the mid-points of sides AB and CD.

Concept Used:

Opposite sides of a parallelogram are equal and parallel.

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.

Diagram:

Proof:

AB || CD

Then,

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 are 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

Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point f DQ

DP = PQ (1)

AB = CD

Similarly,

In ΔAPB,

E is the mid-point of side AB and EQ || AP

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.