In the figure,$ABC$ is a triangle right-angled at $B$ and $BD \perp AC$. If $AD = 4 \, cm$ and $CD = 5 \, cm$,find $BD$ and $AB$.

(N/A) Given,$\triangle ABC$ in which $\angle B = 90^{\circ}$ and $BD \perp AC$.
Also,$AD = 4 \, cm$ and $CD = 5 \, cm$.
In $\triangle ADB$ and $\triangle CDB$,$\angle ADB = \angle CDB = 90^{\circ}$.
Also,$\angle BAD = \angle DBC$ (since both are equal to $90^{\circ} - \angle C$).
Therefore,$\triangle ADB \sim \triangle CDB$ by $AA$ similarity criterion.
Thus,$\frac{BD}{AD} = \frac{CD}{BD}$.
$\Rightarrow BD^2 = AD \times CD = 4 \times 5 = 20$.
$\Rightarrow BD = \sqrt{20} = 2\sqrt{5} \, cm$.
Now,in right-angled $\triangle ADB$,by Pythagoras theorem:
$AB^2 = AD^2 + BD^2 = 4^2 + (2\sqrt{5})^2 = 16 + 20 = 36$.
$\Rightarrow AB = \sqrt{36} = 6 \, cm$.
Hence,$BD = 2\sqrt{5} \, cm$ and $AB = 6 \, cm$.