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.

There must be an error somewhere. How can 3 lines radiating from a point form the sides of a triangle? And what is meant by "the side congruent to PA"? Are there meant to be two triangles?

I went over the problem today, and it turns out that the wording was fallacious. I've seen the solutions for both of these problems now (the way they were meant to be interpreted) and I no longer need help with them...but the ones in my other thread are much more pressing. :P