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 = 3cm², find the area of ||^gm ABCD.

Answer 1

In ΔADF and ΔECF

We have, 

∠ADF = ∠ECF 

AD = EC 

And, 

∠DFA = ∠CFA 

So, by AAS congruence rule, 

ΔADF ≅ ΔECF 

Area (ΔADF) = Area (ΔECF) 

DF = CF 

BF is a median in ΔBCD 

Area (ΔBCD) = 2 Area (ΔBDF) 

Area (ΔBCD) = 2 x 3 

= 6cm²

Hence, 

Area of parallelogram ABCD = 2 Area (ΔBCD) 

= 2 x 6 

= 12 cm²