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(Δ ADF) = ar(Δ ECF)
(ii) If the area of Δ DFB=3 cm², find the area of ||gm ABCD.

Answer 1

ΔADF  ≅ ΔECF , Area of ||gm ABCD. = 12 cm²

Step-by-step explanation:

BC is produced to E such that CE=BC

AD = BC

=> AD = CE

AD ║ AC => AD ║ AE

Comparing ΔADF  & ΔECF

∠DFA = ∠CFE  ( Vertically opposite angle)

∠DAF = ∠CEF   ( as AD ║ AE )

AD = CE

=> ΔADF  ≅ ΔECF

=> DF = CF

=> DF = DC/2

lets draw BM ⊥ DF or DC

Area of Δ DFB  = (1/2) DF  * BM =  3cm²

=> DF * BM = 6

Area of ||gm ABCD. = DC * BM

= 2 * DF * BM

= 2 * 6

= 12

Area of ||gm ABCD. = 12 cm²

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Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

BC is produced to E such that CE=BC

AD = BC

=> AD = CE

AD ║ AC => AD ║ AE

Comparing ΔADF & ΔECF

∠DFA = ∠CFE ( Vertically opposite angle)

∠DAF = ∠CEF ( as AD ║ AE )

AD = CE

=> ΔADF ≅ ΔECF

=> DF = CF

=> DF = DC/2

lets draw BM ⊥ DF or DC

Area of Δ DFB = (1/2) DF * BM = 3cm²

=> DF * BM = 6

Area of ||gm ABCD. = DC * BM

= 2 * DF * BM

= 2 * 6

= 12

Area of ||gm ABCD. = 12 cm²