Question

In given figure D,E,F are the mid-points of the sides BC,CA and AB respectively of a $△ABC$. Determine the ratio of the areas of $△DEF$ and $△ABC$.

Answer 1

Since D and E are the mid-points of the sides BC and AB respectively of $△ABC$.
Therefore,
$DE||BA$

$\Rightarrow$ $DE||FA$........(i)

Since D and F are mid-points of the sides BC and AB respectively of $△ABC$.

$\therefore$ $DF||CA\Rightarrow DF||AE$.......(ii)

From (i), and (ii), we conclude that AFDE is a parallelogram.

Similarly, BDEF is a parallelogram.

Now, in $△DEF$ and $△ABC$, we have

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

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

So, by AA-similarity criterion, we have

$△DEf\sim △ABC$

$\Rightarrow$ $\frac{Area\left(△DEF\right)}{ARE\left(△ABC\right)}=\frac{{DE}^2}{{AB}^2}=\frac{{\left(1/2AB\right)}^2}{{AB}^2}=\frac{1}{4}$ [$\because DE=\frac{1}{2}AB$]

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