Question

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

(i) Prove that ar (Δ ADF) = ar (Δ ECF)

(ii) If the area of Δ DFB = 3 cm², find the area of ||^gm ABCD.

Answer 1

Given: Here from the given figure we get

(1) ABCD is a parallelogram with base AB,

(2) BC is produced to E such that CE = BC

(3) AE intersects CD at F

(4) Area of ΔDFB = 3 cm

To find:

(a) Area of ΔADF = Area of ΔECF

(b) Area of parallelogram ABCD

Proof: Δ ADF and ΔECF, we can see that

∠ADF = ∠ECF (Alternate angles formed by parallel sides AD and CE)

AD = EC

∠DFA = ∠CFA (Vertically opposite angles)

(ASA condition of congruence)

As

DF = CF

Since DF = CF. So BF is a median in ΔBCD

Since median divides the triangle in to two equal triangles. So

Since .So

Hence Area of parallelogram ABCD

Hence we get the result

(a)

(b)