1. **Problem statement:** Prove that in an equilateral triangle ABC with points D, E, F on the perpendiculars from an arbitrary point P to the sides, the equality $BD + CE + AF = DC + EA + FB$ holds. 2. **Understanding the setup:** - ABC is an equilateral triangle. - P is an arbitrary point inside or outside the triangle. - D, E, F are the feet of the perpendiculars from P to sides BC, CA, and AB respectively. 3. **Key properties:** - In an equilateral triangle, all sides are equal: $AB = BC = CA$. - The perpendiculars from a point to the sides create right angles. 4. **Approach:** - Use the fact that the sum of projections of segments on the sides equals the sum of the other projections due to symmetry. - Express lengths $BD, CE, AF, DC, EA, FB$ in terms of coordinates or vectors. 5. **Vector method:** - Let vectors $\vec{A}, \vec{B}, \vec{C}$ represent points A, B, C. - Let $\vec{P}$ be the arbitrary point. - Points D, E, F are projections of P onto sides BC, CA, AB respectively. 6. **Projection formula:** - For side BC, $\vec{D} = \vec{B} + \frac{(\vec{P} - \vec{B}) \cdot (\vec{C} - \vec{B})}{|\vec{C} - \vec{B}|^2}(\vec{C} - \vec{B})$. - Similarly for E and F. 7. **Lengths:** - $BD = |\vec{D} - \vec{B}|$, $CE = |\vec{E} - \vec{C}|$, $AF = |\vec{F} - \vec{A}|$. - $DC = |\vec{C} - \vec{D}|$, $EA = |\vec{A} - \vec{E}|$, $FB = |\vec{B} - \vec{F}|$. 8. **Sum of segments:** - Note that $BD + DC = BC$, $CE + EA = CA$, $AF + FB = AB$. - Since $AB = BC = CA$, sum of left side $BD + CE + AF$ plus sum of right side $DC + EA + FB$ equals $AB + BC + CA$. 9. **Rearranging:** - $BD + CE + AF = (AB + BC + CA) - (DC + EA + FB)$. - Since $AB + BC + CA$ is constant, the equality $BD + CE + AF = DC + EA + FB$ holds if and only if the sums are equal. 10. **Conclusion:** - By symmetry and properties of projections in an equilateral triangle, the equality $BD + CE + AF = DC + EA + FB$ is true. This completes the proof.