Question

in a parallelogram
AB CD, E and f are
The mid points of sides BC and AD respectively
Show that line segments AE and EC trisect the diagonal BD.

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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.


Pls mark brainliest