
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 circumscribed circle around the isosceles triangle \( ABC \). Given that \( BO = 4\sqrt{2} \), \( BH \) (the altitude of the triangle to the base \( AC \)) is \( 2\sqrt{2} \), and \( BH = HO \). The triangle \( AOB \) is equilateral with \( AB = BC = 4\sqrt{2} \). Since \( AB = 2BH \), the angle \( \angle BAC = 30^\circ \). The inscribed angle \( \angle BDC \) subtends the same arc as \( \angle BAC \), therefore \( \angle BDC = 30^\circ \).

Let \( DC = x \). By the Law of Cosines for triangle \( BDC \), we have: 
\[ BC^2 = DB^2 + DC^2 - 2 \cdot DB \cdot DC \cdot \cos 30^\circ \]

This yields the equation:
\[ (4\sqrt{2})^2 = 8^2 + x^2 - 16 \cdot x \cdot \frac{\sqrt{3}}{2} \]

Solving the quadratic equation \( x^2 - 8\sqrt{3}x + 32 = 0 \), we find the roots:
\[ x_1 = 4\sqrt{3} - 4 \]
\[ x_2 = 4\sqrt{3} + 4 \]

Both solutions are valid. The areas \( S_1 \) and \( S_2 \) are calculated as follows:
\[ S_1 = \frac{1}{2} \cdot DB \cdot DC \cdot \sin 30^\circ = 4(4\sqrt{3} - 4) \cdot \frac{1}{2} = 8(\sqrt{3} - 1) \]
\[ S_2 = \frac{1}{2} \cdot DB \cdot DC \cdot \sin 30^\circ = 4(4\sqrt{3} + 4) \cdot \frac{1}{2} = 8(\sqrt{3} + 1) \]

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