Question

D, E and F are the mid point of sides BC , CA,AB of a triangle ABC
1. BDEF is a parallelogram
2. $ar\left(DEF\right)=\frac{1}{4}ar\left(ABC\right)$
3. $\left(BDEF\right)=\frac{1}{2}ar\left(ABC\right)$

Answer 1

(1) From mid point theorems $DEis|{|}^{lel}$ to $AB\&DE=\frac{1}{2}AB:FB$.
FE is $|{|}^{lel}toBC\&FC=\frac{1}{2}BC:BD.$
$\therefore$ BDEF is a parallelogram

(2) Consider $\Delta AFE\&ABC$.
$\begin{array}{cc}\angle AFE=\angle ABC& \\ \angle AEF=\angle ACB& \end{array}\}\because FE||BC.$
$\therefore \Delta AFE\simeq \Delta ABC$.
$\therefore \frac{AE}{AC}=\frac{AF}{AB}=\frac{FE}{BC}=\frac{1}{2}$,
$\therefore \frac{Ar\left(\Delta AFE\right)}{\left(Ar\right(\Delta ABC)}={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$
Similarly $\frac{Ar\Delta FBD}{Ar\Delta ABC}=\frac{1}{4}$
$\left(Ar\Delta FDC\right)/\left(Ar\Delta ABC\right)=\frac{1}{4}$
But $Ar\Delta DEF=Ar\Delta ABC-[Ar\Delta AFE+\Delta FBD+\Delta EDC]$
$=Ar\Delta ABC\{1-(\frac{1}{4}-\frac{1}{4}+\frac{1}{4})\}=\frac{1}{4}Ar.\Delta ABC$
$\Rightarrow \frac{Ar\Delta DEF}{Ar\Delta ABC}=\frac{1}{4}$
Similarly you can prove (iii) also.