Question

D, E, and F are the mid-points of the sides BC, CA and AB, respectively of a $\Delta ABC$. Then find the ratio of the areas of $\Delta DEF$ and $\Delta ABC$.

Answer 1

D and E are the mid-points of the sides BC and AB respectively of $\Delta ABC$.

$\therefore DE\parallel BA,DE=\frac{1}{2}\times BA$ [ Mid-point theorem]

$\Rightarrow DE\parallel FA\dots \dots \left(1\right)$

Since, D and F are mid-points of the sides BC and AB respectively of $\Delta ABC$,

$\therefore DF\parallel CA\Rightarrow DF\parallel AE....\left(2\right)$

From (1) and (2) we conclude that AFDE is a parallelogram

Similarly, BDEF is a parallelogram

Now, in $\Delta DEF$ and $\Delta ABC$, we have

$\angle FDE=\angle A$ [Opposite angles of parallelogram AFDE]

and $\angle DEF=\angle B$ [Opposite angles of parallelogram BDEF]

$\Rightarrow \Delta DEF\sim \Delta ABC$ [AA-similarity criterion]

$\Rightarrow \frac{Area\left(\Delta DEF\right)}{Area\left(\Delta ABC\right)}=\frac{D{E}^2}{A{B}^2}$

$\Rightarrow \frac{Area\left(\Delta DEF\right)}{Area\left(\Delta ABC\right)}=\frac{(\frac{1}{2}AB{)}^2}{A{B}^2}=\frac{1}{4}$

$[\because DE=\frac{1}{2}\times AB]$

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