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 cm2] Thus, ar(||gm ABCD) = 2 ar(△BDC) = 2 × 6 = 12 cm2