**ABCD is a parallelogram in which BC is produced to E such that CE = BC (figure). AE intersects … 3 cm2 , find the area of the…**

###### Answers:

In △ADF and △ECF , we have

∠ADF = ∠ECF [alt.int.∠s]

AD = EC [∵ AD = BC and BC = EC]

∠DFA = ∠CFE [vert. opp. ∠s]

∴ By AAS congruence rule ,

△ADF ≅ △ECF

⇒ DF = CF [c.p.c.t.]

⇒ ar(△ADF) = ar(△ECF)

Now, DF = CF

⇒ BF is a median in △BDC

⇒ ar(△BDC) = 2 ar(△DFB)

= 2 × 3 = 6 cm2 [∵ar(△DFB) = 3 cm^{2}]

Thus, ar(||gm ABCD) = 2 ar(△BDC)

= 2 × 6 = 12 cm^{2}

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