The two parallelogram shaped regions ABCD and AEFG of which ∠A is common are equal in area and E lies on AB. Let’s prove that DE || FC.

Given.

2 parallelogram ABCD and AEFG equal in area with common ∠ A

Formula used.

If 2 triangles are on same base And having equal area then they lies between 2 parallel lines.

If both pair of opposite sides parallel then quadrilateral is parallelogram.

Join DF and EC

Mark intersection point of both parallelogram as O

In quadrilateral AEOD

AE || DO ∵ AB || CD [opposite sides of parallelogram]

AD || EO ∵ AG || GH [opposite sides of parallelogram]

∴ AEOD is a parallelogram

If ABCD is a parallelogram and AEOD is also a parallelogram then EBCO is also a parallelogram

If AEFG is a parallelogram and AEOD is also a parallelogram then DOFG is also a parallelogram

Parallelogram ABCD = Parallelogram AEFG

Parallelogram [AEOD+EBCO]=Parallelogram [AEOD+DOFG]

Cutting both sides parallelogram AEOD

We get ;

Parallelogram EBCO = Parallelogram DOFG

As DF , EC are diagonals of parallelogram EBCO and DOFG

⇒ Diagonal of parallelogram bisect in 2 equal triangles

Hence we get

2× triangle EOC = 2× triangle DOF

Triangle EOC = triangle DOF

Add triangle FOC on both side

Triangle EOC + triangle FOC = triangle DOF + triangle FOC

Triangle ECF= triangle DFC

2 triangles triangle ECF, triangle DFC are equal

And they are on same base FC

Hence they are between parallel lines FC and DE

∴ FC || DE