ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD atF. If the area ofDFB is27 cm . Find the area of parallelogram ABCD.

In a parallelogram, opposite sides and angles are equal. Also, the diagonals of a parallelogram bisect each other.

Given that ABCD is a parallelogram and BC is produced to E such that CE = BC. AE intersects CD at F.

Since CE = BC and ∠B = ∠C (opposite angles of a parallelogram are equal), we can say that triangle BCE is an isosceles triangle.

Therefore, ∠BCE = ∠CBE.

Also, ∠BAC = ∠BCD (alternate angles are equal as AB || CD).

So, ∠BAC = ∠BCE.

This implies that triangle ACF is similar to triangle ECF (by AA similarity criterion).

Therefore, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

Area(ΔACF)/Area(ΔECF) = (AC/EC)^2

But, AC = EC (diagonals of a parallelogram bisect each other).

So, Area(ΔACF) = Area(ΔECF)

Given that Area(ΔDFB) = 27 cm^2, and since ΔDFB and ΔACF are on the same base DF and between the same parallels DF and AC,

Area(ΔACF) = Area(ΔDFB) = 27 cm^2

Since the area of the parallelogram is twice the area of one of its triangles,

Area of parallelogram ABCD = 2 * Area(ΔACF) = 2 * 27 cm^2 = 54 cm^2.

So, the area of parallelogram ABCD is 54 cm^2.