Question

In the figure $△ABC,D,E,F$ are the midpoint of sides $BC,CA$ and $AB$ respectively. Show that
$ar\left(BDEF\right)=\frac{1}{2}ar\left(△ABC\right)$

Answer 1

Given:- $△ABC$ in which $D,E$ and $F$ are the mid-points of sides $BC,CA$ and $AB$ respectively.

To prove:- $ar\left(BDEF\right)=\frac{1}{2}ar\left(△ABC\right)$

Proof:-

We know that line segments joining the mid-point of two sides of a triangle is parallel to the third side.

Now, in $△ABC$,

$F$ is mid-point of $AB$.

$E$ is mid-point of $AC$.

$\therefore EF\parallel BC$

$\Rightarrow EF\parallel BD(\because \text{Parts of parallel lines are parallel})$

Similarly,

$DE\parallel BF$

Now, in $BDEF$ both pairs of opposite sides are parallel.

$\therefore BDEF$ is a parallelogram.

Similarly,

$FDCE$ is a parallelogram.

$AFDE$ is a parallelogram.

As we know that diagonals of a parallelogram divides it into two congruent triangles.

Therefore,

In parallelogram $BDEF$,

$△DBF\cong △DEF$

$\therefore ar\left(△DBF\right)=ar\left(△DEF\right).....\left(1\right)(\because \text{Area of congruent triangles are equal})$

Similarly,

$ar\left(△DEC\right)=ar\left(△DEF\right).....\left(2\right)$

$ar\left(△AFE\right)=ar\left(△DEF\right).....\left(3\right)$

From equation $\left(1\right),\left(2\right)\&\left(3\right)$, we get

$ar\left(△DBF\right)=ar\left(△AFE\right)=ar\left(△DEC\right)=ar\left(△DEF\right)$

Now,

$ar\left(△DBF\right)+ar\left(△AFE\right)+ar\left(△DEC\right)+ar\left(△DEF\right)=ar\left(△ABC\right)$

$\Rightarrow ar\left(△DBF\right)+ar\left(△DBF\right)+ar\left(△DEF\right)+ar\left(△DEF\right)=ar\left(△ABC\right)$

$\Rightarrow 2ar\left(△DBF\right)+2ar\left(△DEF\right)=ar\left(△ABC\right)$

$\Rightarrow 2(ar\left(△DBF\right)+ar\left(△DEF\right))=ar\left(△ABC\right)$

$\Rightarrow 2ar\left(BDEF\right)=ar\left(△ABC\right)$

$\Rightarrow ar\left(BDEF\right)=\frac{1}{2}ar\left(△ABC\right)$

Hence proved.