Question

$D,E\mathrm{and}F$are respectively the mid-points of sides $AB,BC\mathrm{and}CA$ of $\Delta ABC$. Find the ratio of the area of $\Delta DEF$ and $\Delta ABC$

Answer 1

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

In $\Delta ABC$, $F$ is the mid-point of $AB$ and $E$ is the mid-point of $AC$

So, by the mid-point theorem,

$FE\left|\right|BC$ and $FE=\frac{1}{2}\times BC$

$\Rightarrow$$FE\left|\right|BC$ and $FE\left|\right|BD$ …………………….. [$\because$$BD=\frac{1}{2}\times BC$]

Also, we know that, $FE=\frac{1}{2}\times BC$.

Thus, $FE=BD$

As we know, if one pair of opposite sides of a quadrilateral are equal and parallel then it is a parallelogram.

$\therefore$ $BDEF$ is parallelogram.

Step II: Prove that $\Delta FBD$ and $\Delta DEF$ are congruent.

In $\Delta FBD$ and $\Delta DEF$,

$FB=DE$ (Opposite sides of parallelogram $BDEF$)

$FD=FD$ (Common sides)

$BD=FE$ (Opposite sides of parallelogram $BDEF$)

$\therefore$ $\Delta FBD\cong \Delta DEF$

Similarly, we can prove that,

$\Delta AFE\cong \Delta DEF$ and $\Delta EDC\cong \Delta DEF$.

Step III: Comparing the areas of the triangles.

We know that if triangles are congruent, then they are equal in area.

So, $Area\left(\Delta FBD\right)=Area\left(\Delta DEF\right)$ ……………………………$\left(i\right)$

$Area\left(\Delta AFE\right)=Area\left(\Delta DEF\right)$ …………………………….$\left(ii\right)$

$Area\left(\Delta EDC\right)=Area\left(\Delta DEF\right)$ …………………………….$\left(iii\right)$

Also, $Area\left(\Delta ABC\right)=Area\left(\Delta FBD\right)+Area\left(\Delta DEF\right)+Area\left(\Delta AFE\right)+Area\left(\Delta EDC\right)$ ……….…$\left(iv\right)$

Therefore, $Area\left(\Delta ABC\right)=Area\left(\Delta DEF\right)+Area\left(\Delta DEF\right)+Area\left(\Delta DEF\right)+Area\left(\Delta DEF\right)$……………(From $\left(i\right)$, $\left(ii\right)$, $\left(iii\right)$)

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

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

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

So, $Area\left(\Delta DEF\right):Area\left(\Delta ABC\right)=1:4$.

Therefore, the ratio of the area of $\Delta DEF$ and $\Delta ABC$ is $1:4$ .