**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 cm^{2}, 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 $=2 \operatorname{ar}(\triangle B C D)=2 \times 6 \mathrm{~cm}^{2}=12 \mathrm{~cm}^{2}$

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