The figure $ABCDEF$ is a regular hexagon. Find all points $M$ belonging to the hexagon such that 

Area of triangle $MAC =$ Area of triangle $MCD$.
1. **Understanding the Problem:**
   We are given a regular hexagon \(ABCDEF\) and need to find all points \(M\) within the hexagon such that the area of triangle \(MAC\) is equal to the area of triangle \(MCD\).

2. **Properties of a Regular Hexagon:**
   A regular hexagon can be divided into 6 equilateral triangles. Each side of the hexagon is of equal length, and the internal angles are all \(120^\circ\).

3. **Using Symmetry:**
   Since the hexagon is regular, it is symmetric about its center. This symmetry can be used to simplify the problem. We can place the hexagon in a coordinate system with the center at the origin \((0,0)\).

4. **Coordinate System:**
   Let the side length of the hexagon be \(s\). The vertices of the hexagon can be placed at:
   \[
   A = \left(s, 0\right), \quad B = \left(\frac{s}{2}, \frac{s\sqrt{3}}{2}\right), \quad C = \left(-\frac{s}{2}, \frac{s\sqrt{3}}{2}\right), \quad D = \left(-s, 0\right), \quad E = \left(-\frac{s}{2}, -\frac{s\sqrt{3}}{2}\right), \quad F = \left(\frac{s}{2}, -\frac{s\sqrt{3}}{2}\right)
   \]

5. **Area Calculation:**
   The area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
   \[
   \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
   \]

6. **Equating Areas:**
   We need to find \(M = (x, y)\) such that:
   \[
   \text{Area of } \triangle MAC = \text{Area of } \triangle MCD
   \]
   Using the area formula, we get:
   \[
   \frac{1}{2} \left| x(y_C - 0) + s\left(\frac{s\sqrt{3}}{2} - y\right) + \left(-\frac{s}{2}\right)(0 - y) \right| = \frac{1}{2} \left| x(0 - y_D) + \left(-\frac{s}{2}\right)\left(0 - y\right) + \left(-s\right)\left(y - 0\right) \right|
   \]
   Simplifying, we get:
   \[
   \left| x\left(\frac{s\sqrt{3}}{2}\right) + s\left(\frac{s\sqrt{3}}{2} - y\right) - \frac{sy}{2} \right| = \left| -xy + \frac{sy}{2} - sy \right|
   \]

7. **Solving the Equation:**
   Simplifying further, we get:
   \[
   \left| \frac{xs\sqrt{3}}{2} + \frac{s^2\sqrt{3}}{2} - sy - \frac{sy}{2} \right| = \left| -xy - \frac{sy}{2} - sy \right|
   \]
   This equation simplifies to:
   \[
   \left| \frac{xs\sqrt{3}}{2} + \frac{s^2\sqrt{3}}{2} - \frac{3sy}{2} \right| = \left| -xy - \frac{3sy}{2} \right|
   \]

8. **Conclusion:**
   By symmetry and the properties of the hexagon, the line \(CF\) (which passes through the center of the hexagon) is the locus of all points \(M\) such that the areas of \(\triangle MAC\) and \(\triangle MCD\) are equal.

The final answer is the line \( \boxed{ CF } \).