Let $ ABC$ be an arbitrary acute angled triangle. For any point $ P$ lying within the triangle, let
$ D$, $ E$, $ F$ denote the feet of the perpendiculars from $ P$ onto the sides $ AB$, $ BC$, $ CA$ respectively.
Determine the set of all possible positions of the point $ P$ for which the triangle $ DEF$ is isosceles.
For which position of $ P$ will the triangle $ DEF$ become equilateral?
1. **Concyclic Points and Lengths of Perpendiculars:**
   - Since points \( B, D, E, P \) are concyclic, we have:
     \[
     DE = PB \sin B
     \]
   - Similarly, due to the concyclicity of \( A, P, D, F \), we have:
     \[
     DF = PA \sin A
     \]

2. **Condition for \( \triangle DEF \) to be Isosceles:**
   - For \( \triangle DEF \) to be isosceles, two sides must be equal. Suppose \( DE = DF \), then:
     \[
     PB \sin B = PA \sin A
     \]
   - Using the Law of Sines in \( \triangle ABC \), we know:
     \[
     \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}
     \]
   - Therefore, the condition \( PB \sin B = PA \sin A \) can be rewritten as:
     \[
     \frac{PA}{PB} = \frac{\sin B}{\sin A} = \frac{b}{a}
     \]
   - This implies that \( P \) lies on the Apollonius circle corresponding to \( A \) and \( B \).

3. **General Condition for \( \triangle DEF \) to be Isosceles:**
   - Similarly, if \( DE = EF \), then:
     \[
     PB \sin B = PC \sin C
     \]
     which implies:
     \[
     \frac{PC}{PB} = \frac{\sin B}{\sin C} = \frac{b}{c}
     \]
     and \( P \) lies on the Apollonius circle corresponding to \( B \) and \( C \).
   - If \( DF = EF \), then:
     \[
     PA \sin A = PC \sin C
     \]
     which implies:
     \[
     \frac{PC}{PA} = \frac{\sin A}{\sin C} = \frac{a}{c}
     \]
     and \( P \) lies on the Apollonius circle corresponding to \( A \) and \( C \).

4. **Condition for \( \triangle DEF \) to be Equilateral:**
   - For \( \triangle DEF \) to be equilateral, \( P \) must be a common point for two of the three Apollonius circles mentioned above.
   - The only point that satisfies this condition in an acute-angled triangle is the Fermat point (or Torricelli point), which is the point such that the total distance from the vertices of the triangle is minimized and each angle formed by lines connecting this point to the vertices is \( 120^\circ \).

The final answer is the set of points \( \boxed{ P } \) that lie on any of the Apollonius circles corresponding to pairs of vertices \( (A, B) \), \( (B, C) \), or \( (C, A) \) for \( \triangle DEF \) to be isosceles. For \( \triangle DEF \) to be equilateral, \( P \) must be the Fermat point of \( \triangle ABC \).