In the given figure, $\triangle \mathrm{ABC}$ is right angled at $\mathrm{C}$ and $\mathrm{DE} \perp \mathrm{AB}$. Prove that $\triangle \mathrm{ABC} \sim \triangle \mathrm{ADE}$ and hence find the length of $\mathrm{AE}$ and $\mathrm{DE}$.
It is given that $A C B$ is right angle triangle and $\angle C=90^{\circ}$
We have to prove that $\triangle A B C \sim \triangle A D E$ and find the lengths of $A E$ and $D E$.
In $\triangle A B C \sim \triangle A D E$
$\angle \mathrm{A}=\angle \mathrm{A} \quad$ Common $\angle \mathrm{C}=\angle \mathrm{E} \quad 90^{\circ}$
So by $A A$ similarly criterion, we have
$\triangle A B C \sim \triangle A D E$
$\Rightarrow \frac{A B}{A D}=\frac{B C}{D E}=\frac{A C}{A E}$
$\Rightarrow \frac{13 \mathrm{~cm}}{3 \mathrm{~cm}}=\frac{12 \mathrm{~cm}}{D E}=\frac{5 \mathrm{~cm}}{A E}$
Since
$A B^{2}=A C^{2}+B C^{2}$
$=5^{2}+12^{2}$
$=13^{2}$
So $D E=\frac{36}{13} \mathrm{~cm}$
And $A E=\frac{15}{13} \mathrm{~cm}$
Hence, $D E=\frac{36}{13} \mathrm{~cm}$ and $A E=\frac{15}{13} \mathrm{~cm}$