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.
Solution:
In triangles ADF and ECF, we have
∠ADF = ∠ECF [Alternate 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 ≅ ΔECF
⇒ ar(ΔADF) = ar(ΔECF) and DF = CF .
Now, DF = CF
⇒ BF is a median in Δ BCD.
⇒ ar(ΔBCD) = 2ar(ΔBDF)
$\Rightarrow \operatorname{ar}(\Delta B C D)=2 \times 3 \mathrm{~cm}^{2}=6 \mathrm{~cm}^{2}$
Hence, area of a parallelogram $=2 \operatorname{ar}(\triangle B C D)=2 \times 6 \mathrm{~cm}^{2}=12 \mathrm{~cm}^{2}$