In given figure, ABC is a triangle right angled

Solution:

Given, $\triangle A B C$ in which $\angle B=90^{\circ}$ and $B D \perp A C$

Also, $A D=4 \mathrm{~cm}$ and $C D=5 \mathrm{~cm}$

In $\triangle A D B$ and $\triangle C D B, \quad \angle A D B=\angle C D B$[each equal to $90^{\circ}$ ]

and $\quad \angle B A D=\angle D B C \quad$ [each equal to $90^{\circ}-\angle C$ ]

$\therefore \quad \Delta D B A \sim \triangle D C B \quad$ [by AAA similarity criterion]

Then, $\frac{D B}{D A}=\frac{D C}{D B}$

$\Rightarrow \quad D B^{2}=D A \times D C$

$\Rightarrow \quad D B^{2}=4 \times 5$

$\Rightarrow \quad D B=2 \sqrt{5} \mathrm{~cm}$

In right angled $\triangle B D C, \quad B C^{2}=B D^{2}+C D^{2} \quad$ [by Pythagoras theorem]

$\Rightarrow \quad B C^{2}=(2 \sqrt{5})^{2}+(5)^{2}$ 

$=20+25=45$

$\Rightarrow \quad B C=\sqrt{45}=3 \sqrt{5}$

Again, $\triangle D B A \sim \triangle D C B$,

$\frac{D B}{D C}=\frac{B A}{B C}$

$\Rightarrow \quad \frac{2 \sqrt{5}}{5}=\frac{B A}{3 \sqrt{5}}$

$\therefore$ $B A=\frac{2 \sqrt{5} \times 3 \sqrt{5}}{5}=6 \mathrm{~cm}$

Hence, $B D=2 \sqrt{5} \mathrm{~cm}$ and $A B=6 \mathrm{~cm}$