Question

Question 5 (iii)
D, E and F are respectively the mid-points of the sides BC, CA and AB of triangle ABC Show that:
$\left(iii\right)area\left(BDEF\right)=\frac{1}{2}area\left(ABC\right)$

Answer 1

F ,E are the midpoints of the sides AB and AC respectively.
$\therefore FE||BCandFE=\frac{1}{2}BC$
$\Rightarrow FE||BCandFE=BD\left[DismidpointofBC\right].....\left(1\right)$
Similarly D, E are the midpoints of BC and AC respectively.
$\therefore ED||FBandED=\frac{1}{2}AB$
$\Rightarrow ED||FBandED=FB\left[FismidpointofAB\right].....\left(2\right)$
From (1) and (2) BDEF is a parallelogram.
We know that,
area(ABC) = 4 area(DEF)
$=4area\left(\frac{1}{2}areaBDEF\right)$ [For a parallelogram BDEF, diagonal FD divides it into two triangles of equal area]
= 2 area(BDEF)
i.e. area $\left(BDEF\right)=\frac{1}{2}area\left(ABC\right)$