
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

1) Given the angles:
\[
\angle APB = \angle BAC, \quad \angle APB = \angle AKC, \quad \angle AKC = \angle BAC, \quad \angle KAC = \angle ABC.
\]
Segment \( AC \) is a tangent to the circle.

If \( \triangle ABC \approx \triangle AKC \), then:
\[
\frac{AB}{AK} = \frac{AC}{KC} = \frac{BC}{AC} \Rightarrow \frac{AB}{4} = \frac{AC}{3} = \frac{12}{AC} \Rightarrow AC = 6, \quad AB = 8.
\]

2) Let \( CD \) be a median. By the cosine theorem for triangles \( ADC \) and \( BDC \):
\[
AC^2 = AD^2 + CD^2 - 2AD \cdot CD \cos \angle ADC, \quad BC^2 = BD^2 + CD^2 + 2BD \cdot CD \cos \angle ADC.
\]
Since \( AD = BD \), then:
\[
AC^2 + BC^2 = 2AD^2 + 2CD^2,
\]
\[
CD^2 = \frac{1}{2}(AC^2 + BC^2) - AD^2 = \frac{1}{2}(36 + 144) - 16 = 74, \quad CD = \sqrt{74}.
\]

3) Let \( DP = x \), \( DN = y \) (where \( N \) is the intersection of line \( CD \) with the circle, \( N \neq P \)).

The quadrilateral \( ANBP \) is inscribed in a circle \( \Rightarrow AD \cdot DB = PD \cdot BT, \quad 16 = xy \).

By the properties of tangents and secants to the circle:
\[
CN \cdot CP = AC^2, \quad (CD - y) \cdot (CD + x) = AC^2,
\]
\[
(\sqrt{74} - y) \cdot (\sqrt{74} + x) = 36.
\]

4) Solving the system of equations:
\[
16 = xy, \quad (\sqrt{74} - y) \cdot (\sqrt{74} + x) = 36 \Rightarrow y = \frac{22}{\sqrt{74}} + x \Rightarrow x^2 + \frac{22}{\sqrt{74}}x - 16 = 0, \quad x = \frac{-11 + 3\sqrt{145}}{\sqrt{74}}, \quad DP = \frac{-11 + 3\sqrt{145}}{\sqrt{74}}.
\]

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