A point moves so that the sum of the squares of the perpendiculars let fall from it on the sides of an equilateral triangle is constant. Prove that its locus is a circle.

Explanation

To prove that the locus of a point \( P \) moving such that the sum of the squares of the perpendiculars dropped from it onto the sides of an equilateral triangle is constant, we can follow these steps:

### Step 1: Set Up the Triangle
Let the vertices of the equilateral triangle \( ABC \) be defined as follows:
- \( A(-a, 0) \)
- \( B(0, a\sqrt{3}) \)
- \( C(a, 0) \)