Question

ABCD and ABEF are parallelograms.

If O is the mid-point of BC, show that: DC = CF = FE.​

Answer 1

Answer:

consider this fig

Given that ABEF is a rectangle and ABCD is a parallelogram, with the same base AB and have equal areas.

Since ∥gm & rectangle have equal areas with same base AB, thereore it will be between same set of ∥al lines.

AB=CD⟶(1) [opposite sides of the ∥gm]

AB=EF⟶(2) [opposite sides of rectangle]

from (1) & (2), we get

CD=EF⟶(3)

In △ADF,

∠AFD=90

0

AD>AF [∵ Hypotenuse is greater than other sides.] ⟶(4)

BC>BE [Since BC=AD and BE=AF] ⟶(5)

Perimeter of ∥gm ABCD =AB+BC+CD+DA

=AB+BC+EF+DA [from equation (3)]

>EF+FA+AB+BE [Using (5)]

which is the perimeter of rectangle ABEF.

Therefore perimeter of ∥gm with the same base and with equal areas is greater than the perimeter of the rectangle.

∴ Perimeter ABCD > Perimeter ABEF