Sure! Let's solve these problems step by step. I'll provide clear solutions to each question:

4) Triangle $ABC$ with $\angle BAC = 90^\circ$ and $AD \perp BC$:

(i) Prove that $\triangle ADB \sim \triangle CAB$:

1. $\triangle ADB$ and $\triangle CAB$ both share $\angle A$.
2. $\angle ADB = 90^\circ$ (given) and $\angle BAC = 90^\circ$.
3. By the AA similarity criterion (two corresponding angles are equal), $\triangle ADB \sim \triangle CAB$.

(ii) If $BD = 16 \, \text{cm}$, find $AB$:

1. In $\triangle ABC$, $\angle BAC = 90^\circ$, so $AB$ is the hypotenuse.

2. Since $AD$ is perpendicular to $BC$, the triangle $\triangle ADB \sim \triangle CAB$.

   Using similarity, we know: $\frac{AB}{BD} = \frac{BC}{AB}.$

   Let $AB = x$ and $BC = 16 \, \text{cm}$: $x^2 = BD \cdot BC = 16 \cdot x.$

   Solving for $x$: $x = \sqrt{16}.$

**(iii). Find ( Ratio