Question 29
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,
$$
\frac{PD + PE + PF}{AB + BC + CA} = \frac{1}{2\sqrt{3}}
$$
$$
\frac{PD + PE + PF}{AB + BC + CA} = \frac{1}{2\sqrt{3}}
$$