# 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)   $\small ar(DEF)=\frac{1}{4}ar(ABC)$

M manish

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($\Delta$BDF) = Ar ($\Delta$DEF) .............(i)

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

and, Ar($\Delta$AEF) = Ar ($\Delta$DEF)...........(iii)

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

Ar($\Delta$BDF) =  Ar(DCE) = Ar($\Delta$AEF) =  Ar ($\Delta$DEF)

Thus, Ar ($\Delta$ABC) =  Ar($\Delta$BDF) +  Ar(DCE) + Ar($\Delta$AEF) +  Ar ($\Delta$DEF)
Ar ($\Delta$ABC) = 4 . Ar($\Delta$DEF)
$\Rightarrow$ $ar(\Delta DEF) = \frac{1}{4}ar(\Delta ABC)$

Hence proved.

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