CMO 1969 Problem 4

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, $$\frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}.$$