# Q: 5    D, E and F are respectively the mid-points of the sides BC, CA and AB of a . Show that            (ii) We already proved that BDEF is a ||gm.
Similarly, DCEF and DEAF are also parallelograms.
Now, ||gm BDEF and ||gm DCEF is on the same base EF and between same parallels BC and EF
Ar (BDEF) = Ar (DCEF)
Ar(BDF) = Ar (DEF) .............(i)

It is known that diagonals of ||gm divides it into two triangles of equal area.
Ar(DCE) = Ar (DEF).......(ii)

and, Ar(AEF) = Ar (DEF)...........(iii)

From equation(i), (ii) and (iii), we get

Ar(BDF) =  Ar(DCE) = Ar(AEF) =  Ar (DEF)

Thus, Ar (ABC) =  Ar(BDF) +  Ar(DCE) + Ar(AEF) +  Ar (DEF)
Ar (ABC) = 4 . Ar(DEF)

Hence proved.

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