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

In $△\mathrm{ACB}, \angle \mathrm{C}=90° \mathrm{and} \mathrm{CD}\perp \mathrm{AB}$. Prove that $\frac{{\mathrm{BC}}^2}{{\mathrm{AC}}^2}=\frac{\mathrm{BD}}{\mathrm{AD}}$.

In the given figure if $\mathrm{AD}\perp \mathrm{BC}$. Prove that ${\mathrm{AB}}^2+{\mathrm{CD}}^2={\mathrm{BD}}^2+{\mathrm{AC}}^2$

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

$ABC$ is a right triangle right angled at $B$. Let $D$ and $E$ be any points on $AB$ and $BC$ respectively. Prove that ${\mathrm{AE}}^2+{\mathrm{CD}}^2={\mathrm{AC}}^2+{\mathrm{DE}}^2$.