
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
Since \(CC_1\) is the height of triangle \(ABC\) and point \(C_1\) lies on the circle, \(\angle AC_1C = 90^{\circ}\), thus \(AC\) is the diameter. The angle \(\angle AA_1C\) is inscribed and subtends the diameter, hence \(\angle AA_1C = 90^{\circ}\). Therefore, \(\angle AC_1C = \angle AA_1C\).

\(AA_1\) is the median, therefore \(BA_1 = CA_1\). Since \(AA_1\) is both the height and the median of triangle \(ABC\), this triangle is isosceles, thus triangle \(ABC\) is isosceles.

Consider triangle \(BCC_1\), which is a right triangle and \(BA_1 = CA_1\), hence \(C_1A_1\) is the median, and thus \(C_1A_1 = BA_1 = CA_1 = 2\) cm. Therefore, \(BC = 4\) cm.

Consider triangles \(ABA_1\) and \(CBC_1\). In these triangles, \(\angle AA_1B\) and \(\angle CC_1B\) are both \(90^{\circ}\) and \(\angle ABC\) is common. Therefore, these triangles are similar by the AA (Angle-Angle) criterion. Then \(\frac{AB}{BC} = \frac{AA_1}{CC_1} = \frac{3}{2}\). From this, \(AB = 4 \cdot \frac{3}{2} = 6 \, \text{cm} = AC\). Using Heron's formula, we find the area of the triangle: \(S= \sqrt{p(p-a)(p-b)(p-c)}\). Given \(p = \frac{6+6+4}{2} = 8 \, \text{cm}\), then \(S = \sqrt{8 \cdot 4 \cdot 2 \cdot 2} = 8\sqrt{2} \, \text{cm}^2\).

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