# Q: 5  In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. $\small 8.31$). Show that the line segments AF and EC trisect the diagonal BD.

Given: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively

To prove: the line segments AF and EC trisect the diagonal BD.

AB=CD                             (Given)

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

$\Rightarrow AE=CF$         (E and F are midpoints of AB and CD)

AE=CF                             (Given)

AE || CF               (Opposite sides of a parallelogram)

Hence,  AECF is a parallelogram.

In $\triangle$ DCQ,

F is the midpoint of DC.      (given )

FP || CQ           (AECF  is a parallelogram)

By converse of midpoint theorem,

P is the mid point of DQ.

DP= PQ....................1

Similarly,

In $\triangle$ ABP,

E is the midpoint of AB.      (given )

EQ || AP          (AECF  is a parallelogram)

By converse of midpoint theorem,

Q is the midpoint of PB.

OQ= QB....................2

From 1 and 2, we have

DP = PQ = QB.

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

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