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

$△ABC~△DEF$. $BC=3 \mathrm{cm}, EF=4 \mathrm{cm}$ and area of $△ABC=54 {\mathrm{cm}}^2$. Determine the area $△DEF$.

$ABC$ is a triangle and $PQ$ is a straight line meeting $AB$ in $P$ and $AC$ in $Q$. If $AP=1 \mathrm{cm}$ and $BP=3 \mathrm{cm},$ $AQ=1.5 \mathrm{cm}$ and $CQ=4.5 \mathrm{cm}$. Prove that $ar\left(△APQ\right)=\frac{1}{16}\times ar\left(△ABC\right)$.

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

In the given figure, $AD\perp BC$. Prove that $A{B}^2+C{D}^2=B{D}^2+A{C}^2$.