Question

$D,EandF$ are respectively the mid-points of the sides $BC,CAandAB$ of a $\Delta ABC$. Show that:

$\left(i\right)$ $BDEF$ is a parallelogram.

$\left(ii\right)$ $area(∆DEF)$$=$$\frac{1}{4}\times area\left(\Delta ABC\right)$.

$\left(iii\right)$ $area\left(BDEF\right)=\frac{1}{2}\times area\left(\Delta ABC\right)$.

Answer 1

Step I: Prove that $BDEF$ is a parallelogram.

In $\Delta ABC$,

$EF\left|\right|BC$ and $EF=\frac{1}{2}BC$ (by midpoint theorem)

So, $EF\parallel BD$

$BD=\frac{1}{2}BC$ ($D$ is the mid point)

So, $EF=BD$

Also, $BF\mathrm{and}DE$ are parallel and equal to each other.

If one pair of opposite sides are equal and parallel to each other, then it is a parallelogram.

Hence, $BDEF$ is a parallelogram.

Step II: To prove $area(∆DEF)$$=$$\frac{1}{4}\times area\left(\Delta ABC\right)$.

Progressing from the result of $stepI$,

$BDEF,DCEF,AFDE$ are parallelograms.

We know that the diagonal of a parallelogram divides it into two triangles of equal area.

$\therefore$$area\left(\Delta BFD\right)=area\left(\Delta DEF\right)$ ………………….(For parallelogram $BDEF$) — $\left(i\right)$

$area\left(\Delta AFE\right)=area\left(\Delta DEF\right)$………………….. (For parallelogram $DCEF$) — $\left(ii\right)$

$area\left(\Delta CDE\right)=area\left(\Delta DEF\right)$…………………. (For parallelogram $AFDE$) — $\left(iii\right)$

From above equations $\left(i\right),\left(ii\right)\mathrm{and}\left(iii\right)$

$area\left(\Delta BFD\right)=area\left(\Delta AFE\right)=area\left(\Delta CDE\right)=area\left(\Delta DEF\right)$

We know, $area\left(\Delta BFD\right)+area\left(\Delta AFE\right)+area\left(\Delta CDE\right)+area\left(\Delta DEF\right)=area\left(\Delta ABC\right)$

$area\left(\Delta DEF\right)+area\left(\Delta DEF\right)+area\left(\Delta DEF\right)+area\left(\Delta DEF\right)=area\left(\Delta ABC\right)$………(From $\left(i\right),\left(ii\right)\mathrm{and}\left(iii\right)$)

$\Rightarrow$ $4\times area\left(\Delta DEF\right)=area\left(\Delta ABC\right)$

$\Rightarrow$ $area\left(\Delta DEF\right)=\frac{1}{4}\times area\left(\Delta ABC\right)$

Step III: Prove that $area\left(BDEF\right)=\frac{1}{2}\times area\left(\Delta ABC\right)$

We know that the diagonal of a parallelogram divides it into two triangles of equal area.

$Area(paralle\log ramBDEF)=area\left(\Delta DEF\right)+area\left(\Delta BDE\right)$

$\Rightarrow$ $area(paralle\log ramBDEF)=area\left(\Delta DEF\right)+area\left(\Delta DEF\right)$

$\Rightarrow$ $area(paralle\log ramBDEF)=2\times area\left(\Delta DEF\right)$

$\Rightarrow$ $area(paralle\log ramBDEF)=2\times \frac{1}{4}area\left(\Delta ABC\right)$ (From step $\mathrm{II}$)

$\Rightarrow$$area(paralle\log ramBDEF)=\frac{1}{2}\times area\left(\Delta ABC\right)$.

Therefore, it is proved that, $BDEF$ is a parallelogram, $area(∆DEF)$$=$$\frac{1}{4}\times area\left(\Delta ABC\right)$ and $area\left(BDEF\right)=\frac{1}{2}\times area\left(\Delta ABC\right)$.