Problem 481

Let $ABC$ be an equilateral triangle, and $P$ be a point inside this triangle. Let $D$, $E$ and $F$ be the feet of the perpendiculars from $P$ to the sides $BC$, $CA$ and $AB$ respectively. Prove that

a) $AF + BD + CE = AE + BF + CD$ and

b) $[APF] + [BPD] + [CPE] = [APE] + [BPF] + [CPD]$.

The area of triangle $XYZ$ is denoted $[XYZ]$.