Question

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Answer 1

Given, AM and DN are the medians of $\Delta ABCand\Delta DEFrespectively.$
$ToProvethat\frac{area\left(\Delta ABC\right)}{area\left(\Delta DEF\right)}=\frac{A{M}^2}{D{N}^2}$
Proof
In $\Delta ABCand\Delta DEF\text{(Given)}$
$\therefore \frac{area\left(\Delta ABC\right)}{area\left(\Delta DEF\right)}=\frac{A{B}^2}{D{E}^2}...\left(i\right)$
[ Because the areas of two similar triangles are proportional to the squares of their corresponding sides.]
and, $\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}...\left(ii\right)$
$\Rightarrow \frac{AB}{DE}=\frac{\frac{1}{2}BC}{\frac{1}{2}EF}=\frac{CA}{FD}........\left(iii\right)$

In $\Delta ABM\text{and}\Delta DEN$, we have
$\angle B=\angle E$
$\frac{AB}{DE}=\frac{BM}{EN}$ [from (iii)]
$\therefore \Delta ABM\text{and}\Delta DEN$ [By SAS similarity criterion]
$\Rightarrow \frac{AB}{DE}=\frac{AM}{DN}...\left(iii\right)$

$\therefore \frac{area\left(\Delta ABC\right)}{area\left(\Delta DEF\right)}$ $=\frac{A{B}^2}{D{E}^2}=\frac{A{M}^2}{D{N}^2}$.Hence proved.