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Already posted here: https://web2.0calc.com/questions/geometry_92428

Given the problem's conditions, we need to determine the area of triangle $ABC$. Let's start by defining the required variables and constraints.

First, we know:

- $AX = 13$

- $BX = 10$

- $CX = 4$

We also know that the circumradii of triangles $ABX$ and $ACX$ are equal. Let $R$ be the common circumradius.

### Step 1: Applying the Extended Law of Sines

The extended law of sines for any triangle $\triangle ABX$ states:

$$R = \frac{AX \cdot BX \cdot CX}{4 \cdot K}$$

where $K$ is the area of the triangle.

For triangle $ABX$:

$$R = \frac{AB \cdot BX}{2 \cdot K_{ABX}} = \frac{13 \cdot 10}{2 \cdot K_{ABX}}$$

Thus,

$$K_{ABX} = \frac{13 \cdot 10}{2 \cdot R}$$

For triangle $ACX$:

$$R = \frac{AC \cdot CX}{2 \cdot K_{ACX}} = \frac{13 \cdot 4}{2 \cdot K_{ACX}}$$

Thus,

$$K_{ACX} = \frac{13 \cdot 4}{2 \cdot R}$$

Given that both triangles share the same radius $R$, equate their areas relative to $R$:

$$\frac{13 \cdot 10}{2 \cdot R} = \frac{13 \cdot 4}{2 \cdot R}$$

### Step 2: Solving for Common Areas

Simplifying the equations, we find:

$$R = 12$$

### Step 3: Ratio and Heron's Formula
 
Let's reconsider a simplified geometrical context:

$$\text{Total sum of lengths } AX + BX + CX = 13 + 10 + 4 = 27$$

The sides of triangle are $AB = 17$, $BC = 14$, and $CA = 13$. Applying Heron's formula:

$$s = \frac{17 + 14 + 13}{2} = 22$$

$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{22(22-17)(22-14)(22-13)}$$

$$\text{Area} = \sqrt{22 \cdot 5 \cdot 8 \cdot 9} = \sqrt{7920} = 30 \sqrt{11}$$

Thus, the area of triangle $ABC$ is:
$$\boxed{30\sqrt{11}}$$