Equilateral triangle $ABC$ with $P$ inside, $PA= x$, $PB=y$, $PC=z$ and $z^2 =x^2+y^2$. Find side length of $ABC$ 
$ABC$ is an equilateral triangle $ABC$ with $P$ inside it such that $PA= x$, $PB=y$, $PC=z$. If $z^2 =x^2+y^2$ , find the length of the sides of $ABC$ in terms of $x$ and $y$?

If $z^2=x^2+y^2$ then how can I find measures of angles around $P$ so that the sides can be expressed in terms of $x$ and $y$. I've tried everything I can think of.

Well, this is definitely a brute-force method but here's what I can think of:
Draw your triangle with P inside, along with the lines PA, PB, PC. There are 9 angles in total. There are also 4 unknown lengths, $x,y,z$, and the side-length of the equilateral triangle, $L$. In total, this is 13 unknowns.
Let's just call the six outer angles d,e,f,g,h,i and the 3 inner angles (around P) j,k,l.
It's not hard to come up with 13 equations: you know the angles making up the corners of the equilateral triangle add up to 60, i.e. $d+e = 60, f+g = 60, h+i = 60$, and the three on the inside add up to 360, i.e. $j+k+l = 360$. Also, the three "subtriangles" each total angle $180$. So that's 7 equations.
You can also use the sin and cosine rules of triangles incorporating the information about side lengths. This is an additional 3+3 = 6 equations.
In the end, there are 13 unknowns and 13 equations (which do not depend on one another). These can be solved by substitution/elimination to solve for L.
