In given figure, ABC is a triangle right angled at B and BD ⊥ AC. If AD = 4 cm and CD = 5 cm, then find BD and AB.
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 $A B=6 \mathrm{~cm}$