Question

Question 5
D, E and F are respectively the mid-points of sides AB, BC and CA of $\Delta ABC$. Find the ratio of the area of $\Delta DEF$ and $\Delta ABC$.

Answer 1

D and E are the mid-points of $\Delta ABC$
$\therefore$ DE || AC and
$DE=\frac{1}{2}AC\left(Midpointtheorem\right)...\left(i\right)$

In $\Delta BED$ and $\Delta BCA$
$\angle BED=\angle BCA$ (Corresponding angles)
$\angle BDE=\angle BAC$ (Corresponding angles)
$\angle EBD=\angle CBA$ (Common angles)
$\therefore \Delta BED\sim \Delta BCA$ (AAA similarity criterion)

$\frac{ar\left(\Delta BED\right)}{ar\left(\Delta BCA\right)}={\left(\frac{DE}{AC}\right)}^2$
$\Rightarrow \frac{ar\left(\Delta BED\right)}{\Delta BCA}=\frac{1}{4}\left[from\right(i\left)\right]$
$\Rightarrow ar\left(\Delta BED\right)=\frac{1}{4}ar\left(\Delta BCA\right)$
$\Rightarrow ar\left(\Delta BED\right)=\frac{1}{4}ar\left(\Delta BCA\right)$
Similarly, $ar\left(\Delta CFE\right)=\frac{1}{4}ar\left(CBA\right)$ and $ar\left(\Delta ADF\right)=\frac{1}{4}ar\left(\Delta ADF\right)=\frac{1}{4}ar\left(\Delta ABC\right)$
Also, $ar\left(\Delta DEF\right)=ar\left(\Delta ABC\right)-\left[ar\right(\Delta BED)+ar(\Delta CFE)+ar(\Delta ADF\left)\right]$
$\Rightarrow ar\left(\Delta DEF\right)=ar\left(\Delta ABC\right)-\frac{3}{4}ar\left(\Delta ABC\right)=\frac{1}{4}ar\left(\Delta ABC\right)$
$\Rightarrow \frac{ar\left(\Delta DEF\right)}{ar\left(\Delta ABC\right)}=\frac{1}{4}$