<s>[INST] Equilateral Triangle equality 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
 [/INST] 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}
</s>