
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
Let \( O \) be the center of the circumcircle of an isosceles triangle \( A B C \). Then \( B O = 3 \), \( B C \) - the height of the triangle dropped to the base \( A C \), \( B H = 1.5 \), \( B H = H O \), the triangle \( A O B \) is equilateral, \( A B = B C = 3 \). Since \( A B = 2 B H \), the angle \( \angle BAC = 30^{\circ} \). The inscribed angle \( \angle B D C \) subtends the same arc as \( \angle B A C \), thus \( \angle B D C = 30^{\circ} \). Let \( D C = x \). By the law of cosines for triangle \( B D C \), we have:
$$
B C^{2} = D B^{2} + D C^{2} - 2 D B \cdot D C \cos 30^{\circ}
$$
or 
$$
3^{2} = (2\sqrt{3})^{2} + x^{2} - 4\sqrt{3} \cdot x \cdot \frac{\sqrt{3}}{2}
$$
or 
$$
x^{2} - 6x + 3 = 0
$$
$$
x_1 = 3 - \sqrt{6}, \quad x_2 = 3 + \sqrt{6}
$$
Both answers are appropriate.
$$
S_1 = \frac{1}{2} D B \cdot D C \sin 30^{\circ} = \sqrt{3}(3 - \sqrt{6}) \cdot \frac{1}{2} = \frac{3}{2}(\sqrt{3} - \sqrt{2})
$$
$$
S_2 = \frac{1}{2} D B \cdot D C \sin 30^{\circ} = \sqrt{3}(3 + \sqrt{6}) \cdot \frac{1}{2} = \frac{3}{2}(\sqrt{3} + \sqrt{2})
$$

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