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 (ΔADF) = ar (ΔECF)
(ii) If the area of ΔDFB = 3 cm2, find the area of ||^gm ABCD.

Answer 1

In triangles and , ADF ECF we have

∠ADF  = ∠ECF               [Alternative interior angles, Since || AD BE]

AD = EC                        [ Since AD = BC = CE ]

And  ∠DFA  = ∠CFA      [vertically opposite angles]

So, by AAS congruence criterion, we have
ΔADF ≅ ECF

⇒ area (ΔADF)  = area (ΔECF)  and DF = CF.
Now, DF = CF

⇒ BF is a madian in ΔBCD  

⇒ area  (ΔBCD) =2ar  (ΔBDF)

⇒ area (ΔBCD) = 2× 3 cm²  =6cm²

Hence,  ar (||^gm ABCD) = 2ar (ΔBCD) 2 × 6cm²

     =12cm²