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.

The relationship between $x,y,z$ and the side of the triangle, $a$ is interesting:
$$3(x^4+y^4 + z^4 + a^4) = (x^2+y^2+z^2 + a^2)^2 \tag{1}$$
Please see here, near reference $19$
If you put $z^2 = x^2 + y^2$ in $(1)$, you'll get
$$3(x^4 + y^4 + (x^2+y^2)^2 + a^4) = (2x^2 + 2y^2 + a^2)^2$$
Solving for $a$, you get
$$a = \pm \sqrt{x^2+y^2 \pm\sqrt{3}xy}$$
Consider only the positive roots.
