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

STEP 1 : Proving that $\mathbf{△}\mathit{A}\mathit{B}\mathit{M}$ and $\mathbf{△}\mathit{D}\mathit{E}\mathit{N}$ are similar

Let us assume two similar triangles $ABC$ and $DEF$ as shown in figure.

Let $AM$ and $DN$ be the medians of the triangles $ABC$ and $DEF$ respectively.

We know that $△ABC~△DEF$

$\Rightarrow \frac{AB}{DE}=\frac{AC}{DF}=\frac{BC}{EF}$

$\Rightarrow \frac{AB}{DE}=\frac{{\displaystyle \frac{1}{2}}BC}{{\displaystyle \frac{1}{2}}EF}=\frac{BM}{EN}$

Also, $\angle B=\angle E$

Now in $△ABM$ and $△DEN$

$\frac{AB}{DE}=\frac{BM}{EN}$ and $\angle B=\angle E$

$\therefore$By SAS similarity criterion $△ABM~△DEN$

$\Rightarrow \frac{AB}{DE}=\frac{AM}{DN}$ $...\left(1\right)$ [Corresponding sides of similar triangles are proportional]

STEP 2 : Proving that $\frac{\mathbf{a}\mathbf{r}\left(△ABC\right)}{\mathbf{a}\mathbf{r}\left(△DEF\right)}\mathbf{=}{\left(\frac{AM}{DN}\right)}^{\mathbf{2}}$

Since, $△ABC~△DEF$

We know that the areas of two similar triangles are proportional to the squares of the corresponding sides.

$\Rightarrow \frac{ar\left(△ABC\right)}{ar\left(△DEF\right)}={\left(\frac{AB}{DE}\right)}^2$ $...\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get

$\frac{ar\left(△ABC\right)}{ar\left(△DEF\right)}={\left(\frac{AB}{DE}\right)}^2={\left(\frac{AM}{DN}\right)}^2$

$\frac{ar\left(△ABC\right)}{ar\left(△DEF\right)}={\left(\frac{AM}{DN}\right)}^2$

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