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Q: 5    D, E and F are respectively the mid-points of the sides BC, CA and AB of a \small \Delta ABC. Show that
           (ii)   \small ar(DEF)=\frac{1}{4}ar(ABC)

Answers (1)


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 
\therefore  Ar (BDEF) = Ar (DCEF)
\Rightarrow Ar(\DeltaBDF) = Ar (\DeltaDEF) .............(i)

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

and, Ar(\DeltaAEF) = Ar (\DeltaDEF)...........(iii)

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

 Ar(\DeltaBDF) =  Ar(DCE) = Ar(\DeltaAEF) =  Ar (\DeltaDEF)

Thus, Ar (\DeltaABC) =  Ar(\DeltaBDF) +  Ar(DCE) + Ar(\DeltaAEF) +  Ar (\DeltaDEF)
           Ar (\DeltaABC) = 4 . Ar(\DeltaDEF)
         \Rightarrow ar(\Delta DEF) = \frac{1}{4}ar(\Delta ABC)

Hence proved.

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