Question

Question 5 (ii)
D, E and F are respectively the mid-points of the sides BC, CA and AB of triangle ABC Show that:
(ii) $area\left(DEF\right)=\frac{1}{4}area\left(ABC\right)$

Answer 1

F ,E are the midpoints of the sides AB and AC respectively.
$\therefore FE||BCandFE=\frac{1}{2}BC$
$\Rightarrow FE||BCandFE=DC\left[DismidpointofBC\right].....\left(1\right)$
Similarly; D, F are the midpoints of BC and BA respectively.
$\therefore FD||ACandFD=\frac{1}{2}AC$
$\Rightarrow FD||ACandFD=EC\left[EismidpointofAC\right].....\left(2\right)$
From (1) and (2) EFDC is a parallelogram.
$\Rightarrow$ ar(DEF)= ar(DEC) [ diagonal of a parallelogram divides it into two triangles of equal areas] ..... (3)
Similarly, BDFE is a parallelogram
$\Rightarrow$ ar(DEF)= ar(BDF) ...... (4)
and AEDF is a parallelogram
$\Rightarrow$ ar(DEF)= ar(AFE) ...... (5)

Now area(ABC) = area(BDF)+ area(DEF)+ area(DEC)+ area(AEF)
= area(DEF) + area(DEF) + area(DEF) + area(DEF) [From (3), (4) and (5)]
area(ABC) = 4 area(DEF)
i.e. $area\left(DEF\right)=\frac{1}{4}area\left(ABC\right)$