C lies on a circle with diameter AD. B lies on AC and E lies on AD such that BE is parallel to CD. AB=21 cm, CD=18 cm and BE=13.5 cm. (a) Show that triangle ABE is similar to triangle ACD. (b) Find the length of the diameter AD. (c) Work out the circumference of the circle.

Both triangles $\triangle ABE$ and $\triangle ACD$ share the angle at vertex A.

Given that $BE$ is parallel to $CD$ ($BE \parallel CD$).
Since $AC$ is a transversal line intersecting two parallel lines, the corresponding angles are equal.

Alternatively, since $AD$ is a transversal line intersecting two parallel lines, the corresponding angles are equal.

Since two pairs of corresponding angles are equal ($\angle BAE = \angle CAD$ and $\angle ABE = \angle ACD$), the triangles are similar by the Angle-Angle (AA) similarity criterion.
Therefore, $\boxed{\triangle ABE \sim \triangle ACD}$.

14. (b) Find the length of the diameter $AD$.

Since $\triangle ABE \sim \triangle ACD$, the ratio of corresponding sides is equal:

Substitute the given values: $AB = 21 \text{ cm}$, $BE = 13.5 \text{ cm}$, $CD = 18 \text{ cm}$.

To find $AC$, rearrange the equation:

The problem states that C lies on a circle with diameter AD. An angle subtended by a diameter at any point on the circumference is a right angle ($90^\circ$).
Therefore, $\angle ACD = 90^\circ$. This means $\triangle ACD$ is a right-angled triangle with hypotenuse $AD$.

In right-angled $\triangle ACD$:

Substitute the values $AC = 28 \text{ cm}$ and $CD = 18 \text{ cm}$:

The length of the diameter $AD$ is $\boxed{33.3 \text{ cm}}$ (rounded to 3 significant figures).

14. (c) Work out the circumference of the circle.

The circumference $C$ of a circle is given by $C = \pi d$, where $d$ is the diameter.
From part (b), the diameter $d = AD = \sqrt{1108} \text{ cm}$.

Rounding to one decimal place:

The circumference of the circle is $\boxed{104.6 \text{ cm}}$.