ABCD is a parallelogram in which BC is produced to E such that CE = BC and AE intersects CD at F. If ar.(△ DFB) = 30 cm²; find the area of parallelogram.

We know that,

Area of triangles on the same base and between the same parallel lines are equal.

△ ADF and △ DFB lie on same base DF and between same parallel lines AB and DC.

∴ Area of △ ADF = Area of △ DFB = 30 cm²

By converse of mid-point theorem,

If a line is drawn through the midpoint of one side of a triangle, and parallel to the other side, it bisects the third side.

In △ ABE,

C is the mid-point of BE and CF || AB.

∴ F is the mid-point of AE. (By converse of mid-point theorem)

∴ EF = AF.

In △ ADF and △ EFC,

⇒ ∠AFD = ∠EFC (Vertically opposite angles are equal)

⇒ EF = AF (Proved above)

⇒ ∠DAF = ∠CEF (Alternate interior angles are equal)

∴ △ ADF ≅ △ EFC (By A.S.A. axiom)

We know that,

Area of congruent triangles are equal.

∴ Area of △ EFC = Area of △ ADF = 30 cm².

In △ BFE,

Since, C is the mid-point of BE.

∴ CF is the mid-point of median of triangle.

We know that,

Median of triangle divides it into two triangles of equal areas.

∴ Area of △ BFC = Area of △ EFC = 30 cm².

From figure,

⇒ Area of △ BDC = Area of △ BDF + Area of △ BFC

⇒ Area of △ BDC = 30 + 30 = 60 cm².

We know that,

The area of triangle is half that of a parallelogram on the same base and between the same parallels.

From figure,

|| gm ABCD and △ BDC lies on same base DC and between same parallel lines AB and DC.

∴ Area of △ BDC = $\dfrac{1}{2}$ Area of || gm ABCD

⇒ Area of || gm ABCD = 2 × Area of △ BDC

⇒ Area of || gm ABCD = 2 × 60 = 120 cm².

Hence, area of || gm ABCD = 120 cm².