
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
ABCD is an isosceles trapezoid with \(AB = CD\). \(\angle A\) is acute, \(AB\) is the diameter of a circle, \(M\) is the center of the circle, and \(P\) is the point of tangency of the circle with the side \(CD\). Denote the radius of the circle as \(x\). Then \(AM = MB = MP = x\). Let \(N\) be the midpoint of the side \(CD\), making the triangle \(MPN\) right-angled with \(\angle P = 90^{\circ}\), \(\angle MNP = \angle A = \alpha\), and \(MN = \frac{x}{\sin \alpha}\). Let \(K\) be the point where the circle intersects the base \(AD\) (with \(K \neq A\)). Triangle \(ABK\) is right-angled with \(\angle K = 90^{\circ}\), and \(BK\) is the height of the trapezoid with \(BK = AB \sin \alpha = 2x \sin \alpha\).

Given that the area of the trapezoid is \(S_{ABCD} = MN \cdot BK = 2x^2 = 450\), solve for \(x\).

Then, \(AK = AB \cos \alpha = 2x \cos \alpha\) and \(KD = MN = \frac{x}{\sin \alpha}\).

Given that \(\frac{AK}{KD} = \frac{24}{25}\), solve to find the values of \(\sin \alpha\) and \(\cos \alpha\).

For the first case, find \(AD\) and \(BC\).

For the second case, find \(AD\) and \(BC\).

Present the answer in LaTex format: \boxed{Your answer}
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