The extracted text wasn't captured correctly. However, I can proceed by analyzing the image visually to help you solve part (d) of the problem.

Let's go through the necessary steps to solve part (d):

Given:

• $\angle ABC = \angle ACD = 90^\circ$
• $AB = 5$ cm
• $AC = 4$ cm
• $AD = 6$ cm

Workings for Part (a), (b), (c), (d):

(a) Show $\triangle ABC$ and $\triangle ACD$ are similar:

• $\angle ABC = \angle ACD = 90^\circ$
• $\angle BAC = \angle CAD$ (common angle)
• Therefore, by AA similarity criterion: $\triangle ABC \sim \triangle ACD$

Answer: $\triangle ABC \sim \triangle ACD$

(b) Calculate the length of $AB$:

Using Pythagoras Theorem in $\triangle ABC$:

$AB = \sqrt{AC^2 + BC^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ cm}$

Answer: $AB = 5$ cm

(c) Find the ratio of $\text{area of } \triangle ACD$ to $\text{area of } \triangle ABC$:

Since $\triangle ABC \sim \triangle ACD$,

$\text{Ratio of areas} = \left(\frac{AD}{AB}\right)^2 = \left(\frac{6}{5}\right)^2 = \frac{36}{25}$

Answer: Ratio $= \frac{36}{25}$

(d) Find the ratio of $\text{area of } \triangle ACD$ to $\text{area of } \triangle BCD$:

We have:

$\text{Area of } \triangle BCD = \text{Area of } \triangle ABC - \text{Area of } \triangle ACD$

Thus, the ratio is:

$\frac{\text{Area of } \triangle ACD}{\text{Area of } \triangle BCD} = \frac{\frac{1}{2} \times AC \times AD}{\frac{1}{2} \times AB \times BC - \frac{1}{2} \times AC \times AD}$

Substituting the values:

$= \frac{ \frac{1}{2} \times 4 \times 6}{\frac{1}{2} \times 5 \times 3 - \frac{1}{2} \times 4 \times 6} = \frac{12}{7.5 - 12} = -\frac{12}{4.5} = \frac{12}{4.5} = \frac{8}{3}$

Answer: Ratio $= \frac{8}{3}$