1. **Problem statement:** We have an equilateral triangle ABC and a point P inside it. From P, three perpendiculars are dropped to the sides BC, AC, and AB, meeting at points D, E, and F respectively. We need to prove that the sum of the lengths PD + PE + PF is constant, independent of the location of P inside the triangle. 2. **Key fact:** In an equilateral triangle, the sum of the perpendicular distances from any interior point to the three sides is equal to the altitude (height) of the triangle. 3. **Formula for altitude:** For an equilateral triangle with side length $a$, the altitude $h$ is given by $$h = \frac{\sqrt{3}}{2}a$$ 4. **Explanation:** The perpendicular distances PD, PE, and PF from point P to the sides BC, AC, and AB respectively add up to the altitude $h$. 5. **Proof outline:** - Let the perpendicular distances from P to sides BC, AC, and AB be $d_1$, $d_2$, and $d_3$ respectively. - By Viviani's theorem, for any point inside an equilateral triangle, $$d_1 + d_2 + d_3 = h$$ 6. **Conclusion:** Since PD, PE, and PF are these perpendicular distances, their sum is constant and equals the altitude of the triangle, independent of the position of P. **Final answer:** $$PD + PE + PF = \frac{\sqrt{3}}{2}a$$ which is constant for any point P inside the equilateral triangle ABC.