Problem: Point P is on arc AB of the circumcircle of regular hexagon ABCDEF. Prove that PD + PE = PA + PB + PC + PF. I'm aware that I'm supposed to use Ptolemy's theorem, which states that I've drawn the hexagon and it seems like an unreasonably long proof is required if I wanted to prove it based on various quadrilaterals inherent within the circle, but I've got a hunch that I can use the analog for a triangle: which I've already proved in a separate question. Any tips for getting the ball rolling for this proof? Also, there's another question that I have based on using Ptolemy's theorem: For this problem, primarily, I'm having a difficult time determining how to draw the figure. I'm sure that once I am able to, I'll be able to work through to a solution using Ptolemy's inequality, but how should I go about making a figure for this? I find the wording to be a bit abstract compared to how it usually is in these types of problems.