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.

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#### Solution

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 cm^{2 }=6cm^{2}

Hence, ar (||^{gm }ABCD) = 2ar (ΔBCD) 2 × 6cm^{2}

=12cm^{2}

^{}

Concept: Concept of Area

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