Let $ABC$ be an acute triangle. Find the locus of the points $M$, in the interior of $\bigtriangleup ABC$, such that $AB-FG= \frac{MF.AG+MG.BF}{CM}$, where $F$ and $G$ are the feet of the perpendiculars from $M$ to the lines $BC$ and $AC$, respectively.
1. Given the problem, we need to find the locus of points \( M \) inside the acute triangle \( \triangle ABC \) such that the given equation holds:
   \[
   AB - FG = \frac{MF \cdot AG + MG \cdot BF}{CM}
   \]
   where \( F \) and \( G \) are the feet of the perpendiculars from \( M \) to the lines \( BC \) and \( AC \), respectively.

2. Start by rewriting the given equation:
   \[
   (AB)(CM) - (FG)(CM) = (MF)(AG) + (MG)(BF)
   \]
   Using Ptolemy's Theorem in the cyclic quadrilateral \( MGCF \), we have:
   \[
   (AB)(CM) = (MF)(AG + GC) + (MG)(BF + FC)
   \]

3. Simplify the equation:
   \[
   (AB)(CM) = (MF)(AC) + (MG)(BC)
   \]

4. Express \( MF \) and \( MG \) in terms of \( CM \) and the angles:
   \[
   MF = CM \sin \angle MCB \quad \text{and} \quad MG = CM \sin \angle MCA
   \]

5. Substitute these into the equation:
   \[
   (AB)(CM) = (CM \sin \angle MCB)(AC) + (CM \sin \angle MCA)(BC)
   \]

6. Factor out \( CM \):
   \[
   AB = AC \sin \angle MCB + BC \sin \angle MCA
   \]

7. Let \( CD' \) be the isogonal conjugate of \( CM \), where \( D' \) lies on \( AB \). Then:
   \[
   AB = AC \sin \angle ACD' + BC \sin \angle BCD'
   \]

8. Consider the area relationships:
   \[
   \frac{1}{2}(AB)(CD') = \frac{1}{2}(CD')(AC) \sin \angle ACD' + \frac{1}{2}(CD')(BC) \sin \angle BCD'
   \]

9. Simplify the area equation:
   \[
   \frac{1}{2}(AB)(CD') = [ACD'] + [BCD']
   \]

10. Since the total area of \( \triangle ABC \) is:
    \[
    \frac{1}{2}(AB)(CD') = [ABC]
    \]

11. This implies that \( AB \perp CD' \).

12. Therefore, the locus of \( M \) is all points in the interior of \( \triangle ABC \) such that \( M \) lies on \( AO \), where \( O \) is the circumcenter of the triangle. This is because the circumcenter and orthocenter are isogonal conjugates.

\(\blacksquare\)

The final answer is the locus of \( \boxed{ M } \) is all points in the interior of \( \triangle ABC \) such that \( M \) lies on \( AO \), where \( O \) is the circumcenter of the triangle.