Show that ACD is similar to BCD.

To prove similarity, we need to show that two angles in $\triangle ACD$ are equal to two angles in $\triangle BCD$.

• $\angle ACD = \angle BCD$ (Common angle)
• $\angle CAD = \angle BDC$ (Angle between tangent DC and chord DB is equal to the angle in the alternate segment, $\angle CAD$).

Since two angles are equal, the triangles are similar by AA similarity criterion.
Therefore, $\triangle ACD \sim \triangle BCD$.

Given $BC = 4$ cm and $CD = 6$ cm.
Using the Tangent-Secant Theorem, which states that the square of the length of the tangent from an external point to a circle is equal to the product of the lengths of the secant segment and its external segment.

$$CD^2 = BC \times AC$$

Substitute the given values:

$$6^2 = 4 \times AC$$ $$36 = 4 \times AC$$

Divide by 4 to find AC:

$$AC = \frac{36}{4}$$ $$AC = 9 \text{ cm}$$

Since $AC = AB + BC$, we can find AB:

$$AB = AC - BC$$ $$AB = 9 - 4$$ $$AB = 5 \text{ cm}$$

The length of AB is $\boxed{\text{5 cm}}$.

From the similarity $\triangle ACD \sim \triangle BCD$, we have the ratio of corresponding sides:

$$\frac{AC}{CD} = \frac{CD}{BD}$$

Substitute the known values $AC = 9$ cm and $CD = 6$ cm:

$$\frac{9}{6} = \frac{6}{BD}$$

Cross-multiply:

$$9 \times BD = 6 \times 6$$ $$9 \times BD = 36$$ $$BD = \frac{36}{9}$$ $$BD = 4 \text{ cm}$$

Now consider $\triangle BCD$. We have sides $BC = 4$ cm, $CD = 6$ cm, and $BD = 4$ cm.
We can use the Law of Cosines to find $\angle BCD$ (which is $\angle ACD$):

$$BD^2 = BC^2 + CD^2 - 2(BC)(CD)\cos(\angle BCD)$$

Substitute the side lengths:

$$4^2 = 4^2 + 6^2 - 2(4)(6)\cos(\angle BCD)$$ $$16 = 16 + 36 - 48\cos(\angle BCD)$$ $$16 = 52 - 48\cos(\angle BCD)$$

Rearrange to solve for $\cos(\angle BCD)$:

$$48\cos(\angle BCD) = 52 - 16$$ $$48\cos(\angle BCD) = 36$$ $$\cos(\angle BCD) = \frac{36}{48}$$ $$\cos(\angle BCD) = \frac{3}{4}$$

Now, find the angle:

$$\angle BCD = \arccos\left(\frac{3}{4}\right)$$ $$\angle BCD \approx 41.4096^\circ$$

Rounding to one decimal place:

$$\angle ACD \approx 41.4^\circ$$

The size of angle $\angle ACD$ is \boxed{\text{41.4^\circ}}.

The line AOB is a straight line passing through the center O, which means AB is the diameter of the circle.
From Step 2, we found $AB = 5$ cm.
So, the diameter $d = 5$ cm.
The radius $r = \frac{d}{2} = \frac{5}{2} = 2.5$ cm.
The area of a circle is given by the formula $A = \pi r^2$.

$$A = \pi (2.5)^2$$ $$A = \pi (6.25)$$ $$A \approx 19.63495... \text{ cm}^2$$

Rounding to 1 decimal place:

$$A \approx 19.6 \text{ cm}^2$$

The area of the circle is $\boxed{\text{19.6 cm}^2}$.