Question -
Answer -
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 (Alternateangles formed by parallel sides AD and CE)
AD = EC
∠DFA = ∠CFA (Verticallyopposite 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 equaltriangles. So
Since 
.So
Hence Area of parallelogram ABCD
Hence we get the result
(a)
(b)