Question

"Question 5 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.

Class 9 - Math - Quadrilaterals Page 151"

Answer 1

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...

Answer 2

Hi friend ✋

GIVEN:-E and F are the mid point of sides AB and CD of the parallelogram ABCD whose diagonal is BD.

TO PROVE:-BQ=QP=PD

PROOF:-ABCD is parellogram(given)

AB||DC and AB||DC(opposite side is ||gm)

E is the mid point of AB

AE=1/2AB............((1))

F is the mid point CD

CF=1/2CD

CF=1/2AB ------(2)

From 1 and 2

AE =CF.

Also AE || CF

Thus, a pair of opposite sides os a quadrilateral AECF are parallel and equal .

Quadrilateral,AECF id a parellogram

=EC || AF


=EQ || AP and QC || MF
.

In triangle BMA ,E. is the mid point of BA. given

EQ || AP. proved

BQ=LP


Similar by taking triangle CLD ,we can prove that
DP=QP

From 3 and 4 we get

BQ=QP=PD

Hence, AF and CE trisect the diagonal AC.

Hope help ....