Question

17.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.
​

Answer 1

$hello \: frnd$

Parallelogram :

A quadrilateral in

which both pairs of opposite sides are parallel is called a parallelogram.

A

quadrilateral is a parallelogram if

i)Its

opposite sides are equal

ii)

its opposite angles are equal

iii)

diagonals bisect each other

iv)

a pair of opposite sides is equal and parallel.

Converse

of mid point theorem:

The

line drawn through the midpoint of one side of a triangle, parallel to another

side bisect the third side.

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Given,

ABCD is a parallelogram. E and F are the mid-points of sides AB and CD

respectively.

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

Proof,

ABCD is a parallelogram

Therefore, AB || CD

also, AE || FC

Now,

AB = CD

(Opposite sides of parallelogram ABCD)

1/2 AB = 1/2 CD

AE = FC

 (E and F are

midpoints of side AB and CD)

Since a pair of opposite sides of a

quadrilateral AECF is equal and parallel.

so,AECF is a

parallelogram

Then, AF||EC,

AP||EQ & FP||CQ

(Since opposite sides of a

parallelogram are parallel)

Now,

In ΔDQC,

F is mid point of side DC & FP || CQ

(as AF || EC).

So,P is the

mid-point of DQ

 (by Converse of mid-point theorem)

DP = PQ — (i)

Similarly,

In APB,

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

(as AF || EC).

So,Qis the mid-point of PB

(by Converse of mid-point theorem)

 PQ = QB —

(ii)

From equations (i) and (ii),

DP = PQ = BQ

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

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  Hope this will help you...

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