Let $ABC$ be an equilateral triangle. For a point $M$ inside $\vartriangle ABC$, let $D,E,F$ be the feet of the perpendiculars from $M$ onto $BC,CA,AB$, respectively. Find the locus of all such points $M$ for which $\angle FDE$ is a right angle.


1. **Define the coordinates of the vertices of the equilateral triangle:**
   Let the vertices of the equilateral triangle \( \triangle ABC \) be \( A(0, \frac{\sqrt{3}}{2}a) \), \( B(-\frac{a}{2}, 0) \), and \( C(\frac{a}{2}, 0) \).

2. **Define the coordinates of point \( M \):**
   Let \( M(x, y) \) be a point inside \( \triangle ABC \).

3. **Determine the feet of the perpendiculars from \( M \) to the sides of the triangle:**
   - The foot of the perpendicular from \( M \) to \( BC \) is \( D \).
   - The foot of the perpendicular from \( M \) to \( CA \) is \( E \).
   - The foot of the perpendicular from \( M \) to \( AB \) is \( F \).

4. **Use the condition \( \angle FDE = 90^\circ \):**
   Since \( \angle FDE = 90^\circ \), the points \( F, D, E \) form a right angle at \( D \).

5. **Apply the geometric property of the right angle in a circle:**
   The right angle inscribed in a circle implies that \( D \) lies on the circle with diameter \( FE \).

6. **Determine the locus of \( M \):**
   The locus of \( M \) such that \( \angle FDE = 90^\circ \) is a circle. This circle is centered at \( (0, -\frac{\sqrt{3}}{2}a) \) with radius \( a \).

   The equation of this circle is:
   \[
   x^2 + \left(y + \frac{\sqrt{3}}{2}a\right)^2 = a^2
   \]

7. **Verify the locus is inside the triangle:**
   Since \( M \) is inside \( \triangle ABC \), the locus of \( M \) is the part of the circle that lies within the triangle.

The final answer is \( \boxed{ x^2 + \left(y + \frac{\sqrt{3}}{2}a\right)^2 = a^2 } \).