# Prove that the ratio of the areas of two similar triangles is equal

Question.

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

Solution:

In figure, $\mathrm{AD}$ is a median of $\triangle \mathrm{ABC}$ and $\mathrm{PM}$ is a median of $\triangle \mathrm{PQR}$. Here, $\mathrm{D}$ is mid-point of $\mathrm{BC}$ and $M$ is mid-point of $Q R$.

Now, we have $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$.

$\Rightarrow \angle \mathrm{B}=\angle \mathrm{Q}$ ...(1)

(Corresponding angles are equal)

Also $\quad \frac{A B}{P Q}=\frac{B C}{Q R}$

(Ratio of corresponding sides are equal)

$\Rightarrow \frac{A B}{P Q}=\frac{2 B D}{2 Q M}$

( $\because \mathrm{D}$ is mid-point of $\mathrm{BC}$ and $\mathrm{M}$ is mid-point of $\mathrm{QR}$ )

$\Rightarrow \frac{A B}{P O}=\frac{B D}{Q M}$ ...(2)

In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{POM}$

$\angle \mathrm{ABD}=\angle \mathrm{PQM}$ (by1)

and $\frac{A B}{P Q}=\frac{B D}{O M}$(by2)

$\Rightarrow \Delta \mathrm{ABD} \sim \Delta \mathrm{PQM}$ (SAS similarity) $\Rightarrow \frac{A B}{P Q}=\frac{A D}{P M}$ ...(3)

Now, $\frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta P Q R)}=\frac{A B^{2}}{P Q^{2}}$ (SAS similarity)

$\Rightarrow \frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta P O R)}=\frac{A D^{2}}{P M^{2}} \quad\left(\because \frac{A B}{P Q}=\frac{A D}{P M}\right)$