1. Introduction to Barycentric Coordinates:
We start by considering the point $M$ inside triangle $\triangle ABC$ such that the areas of triangles $\triangle ABM$, $\triangle BCM$, and $\triangle CAM$ are equal. We will use Barycentric coordinates with respect to $\triangle ABC$. In Barycentric coordinates, the vertices of $\triangle ABC$ are represented as:
$$A = (1, 0, 0), \quad B = (0, 1, 0), \quad C = (0, 0, 1)$$

2. Equal Area Condition:
For the areas of $\triangle ABM$, $\triangle BCM$, and $\triangle CAM$ to be equal, the point $M$ must divide the triangle into three regions of equal area. This implies that the coordinates of $M$ must be:
$$M = \left( \frac{1}{3}, \frac{1}{3}, \frac{1}{3} \right)$$
This is because the sum of the Barycentric coordinates must be 1, and equal areas imply equal weights.

3. Midpoints of the Sides:
Next, we find the midpoints of the sides of $\triangle ABC$:
$$D = \left( \frac{1}{2}, \frac{1}{2}, 0 \right) \quad \text{(midpoint of } AB\text{)}$$
$$E = \left( 0, \frac{1}{2}, \frac{1}{2} \right) \quad \text{(midpoint of } BC\text{)}$$
$$F = \left( \frac{1}{2}, 0, \frac{1}{2} \right) \quad \text{(midpoint of } CA\text{)}$$

4. Equations of the Medians:
We now find the equations of the medians $AD$, $BE$, and $CF$ in Barycentric coordinates:
- The median $AD$ passes through $A = (1, 0, 0)$ and $D = \left( \frac{1}{2}, \frac{1}{2}, 0 \right)$.
- The median $BE$ passes through $B = (0, 1, 0)$ and $E = \left( 0, \frac{1}{2}, \frac{1}{2} \right)$.
- The median $CF$ passes through $C = (0, 0, 1)$ and $F = \left( \frac{1}{2}, 0, \frac{1}{2} \right)$.

5. Intersection of the Medians:
By Barycentric Ceva's theorem, the medians of a triangle intersect at a single point, which is the centroid. The centroid $G$ of $\triangle ABC$ in Barycentric coordinates is:
$$G = \left( \frac{1}{3}, \frac{1}{3}, \frac{1}{3} \right)$$

6. Conclusion:
Since $M$ has the same Barycentric coordinates as the centroid $G$, we conclude that $M$ is indeed the intersection point of the medians of $\triangle ABC$.

$\blacksquare$