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

Given:- ABCD is a parallelogram

E and F are the midpoints of the sides AB and DC respectively

Formula used:- Line drawn through midpoint of one side of triangle

Parallel to other side , bisect the 3^rd side.

Solution:-

As ABCD is parallelogram

AB=CD and AB||CD;

⇒ AE=CF and AE||CF [E and F are the midpoints of AB and CD]

In quadrilateral AECF

⇒ AE=CF and AE||CF

AECF is a parallelogram.

∵ Quadrilateral having one pair of side equal and parallel are parallelogram

If AECF is a parallelogram

∴ AF||CE

Hence ;

PF||CQ and AP||QE [As AF=AP+PF and CE=CQ+QE ]

In Δ DQC

DF=FC [F is midpoint]

PF||CQ

Then;

P is midpoint of DQ

DP=PQ

∵ Line drawn through midpoint of one side of triangle Parallel to

other side , bisect the 3^rd side

In Δ APB

AE=EB [E is midpoint]

AP||QE

Then;

Q is midpoint of PB

PQ=QB

∵ Line drawn through midpoint of one side of triangle Parallel to

other side , bisect the 3^rd side

If DP=PQ and PQ=QB

Then DP=PQ=QB

Conclusion:-

Line segments AF and EC trisect the diagonal BD.