In the diagram, $AB$ and $CD$ are chords of a circle, with $AB = 9\text{ cm}$ and $CD = 25\text{ cm}$. It is given that $AD$ and $BC$ intersect each other at $90^\circ$ at $F$ and $\angle BCD = \theta$, where $\theta$ varies.

1. Identify angles in the same segment

Angles in the same segment are equal, so $\angle BAD = \angle BCD = \theta$.

2. Express AD as the sum of AF and FD using trigonometry in right-angled triangles

In right-angled $\triangle ABF$, $\cos \angle BAF = \frac{AF}{AB} \implies AF = 9 \cos \theta$.
In right-angled $\triangle CDF$, $\sin \angle FCD = \frac{FD}{CD} \implies FD = 25 \sin \theta$.
Therefore, $AD = AF + FD = 9 \cos \theta + 25 \sin \theta$.

1. Expand $R \cos(\theta - \alpha)$ and compare coefficients with $9 \cos \theta + 25 \sin \theta$

$R \cos(\theta - \alpha) = R \cos \theta \cos \alpha + R \sin \theta \sin \alpha$
Comparing with $9 \cos \theta + 25 \sin \theta$:
$R \cos \alpha = 9$
$R \sin \alpha = 25$

2. Calculate the value of $R$

$R = \sqrt{9^2 + 25^2}$
$R = \sqrt{81 + 625} = \sqrt{706} \approx 26.571$

3. Calculate the value of $\alpha$ and write the final expression

$\tan \alpha = \frac{25}{9}$
$\alpha = \tan^{-1}\left(\frac{25}{9}\right) \approx 70.201^\circ$
$AD = \sqrt{706} \cos(\theta - 70.2^\circ)$

1. Express $AD$ using the $R$-formula result from part (ii)

From part (ii), $R = \sqrt{9^2 + 25^2} = \sqrt{706}$ and $\tan \alpha = \frac{25}{9} \implies \alpha \approx 70.201^\circ$.
So, $AD = \sqrt{706} \cos(\theta - 70.201^\circ)$.

2. Set up the equation for $AD = 26$ and find the basic angle

$\sqrt{706} \cos(\theta - 70.201^\circ) = 26$
$\cos(\theta - 70.201^\circ) = \frac{26}{\sqrt{706}}$
Basic angle $= \cos^{-1}\left(\frac{26}{\sqrt{706}}\right) \approx 11.903^\circ$

3. Solve for the possible values of $\theta$

$\theta - 70.201^\circ = 11.903^\circ \text{ or } -11.903^\circ$
$\theta = 70.201^\circ + 11.903^\circ = 82.104^\circ \approx 82.1^\circ$
$\theta = 70.201^\circ - 11.903^\circ = 58.298^\circ \approx 58.3^\circ$

1. Identify the condition for $AD$ to be the diameter

The diameter is the longest chord in a circle. Therefore, $AD$ is the diameter when its length is at a maximum.

2. Find the angle $\theta$ that maximizes $AD$

From part (ii), $AD = \sqrt{706} \cos(\theta - 70.201^\circ)$.
For $AD$ to be maximum, $\cos(\theta - 70.201^\circ) = 1$.
$\theta - 70.201^\circ = 0^\circ$
$\theta = 70.201^\circ \approx 70.2^\circ$ (to 1 d.p.)