Equilateral Triangle equality

Question

Let ABC be an equilateral triangle, and P be an arbitrary point within the triangle. Perpendiculars PD, PE, PF are drawn to the three sides of the triangle. Show that, no matter where P is chosen, PD + PE + PF / AB + BC + CA = 1/2√3

Answer 1

Let $AB=BC=AC=a$. It follows from the areas equality:

\begin{align} S_{\triangle APB}+S_{\triangle APC}+S_{\triangle BPC} &= S_{\triangle ABC} \\ \frac{1}{2}PD\cdot a+\frac{1}{2}PE\cdot a+\frac{1}{2}PF\cdot a &= a^2\frac{\sqrt{3}}{4} \\ PD+PE+PF &= \frac{3a}{2\sqrt{3}} \\ \frac{PD+PE+PF}{AB+BC+AC} &= \frac{1}{2\sqrt{3}}. \end{align}

Answer 2

the area of triangle ABC is equal to the sum of the areas of the triangles PAB, PBC, PCA. but each of thes is equal to half the product of a perpendicular from P to a side of ABC with the length of that side. since the latter are all equal, the result follows. the numerical constant is easy to estimate using the limiting cases where P approaches a vertex of the triangle ABC