Question

In the given figure, ABCD is a parallelogram and E. Fare the mid-points of sides
AB and CD respectively. Show that the line segments AF and EC trisect the
diagonal D​

Answer 1

Given here is:-

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

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

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

AS WE KNOW

ABCD is a parallelogram

Therefore, AB || CD

also, AE || FC

SO,

=AB = CD

(Opposite sides of parallelogram ABCD)

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

AE = FC

(E and F are the midpoints of side AB and CD)

a pair of opposite sides of a

a pair of opposite sides of aquadrilateral AECF is equal and parallel.

so,AECF is a

parallelogram

Then, AF||EC,

AP||EQ & FP||CQ

( 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, AF and CE trisect the diagonal AC.