Find the side of an equilateral triangle given only the distance of an arbitrary point to its vertices Triangle $ABC$ is an equilateral triangle and $P$ is an arbitrary point inside it. The distance from $P$ to $A$ is $4$ and the distance from $P$ to $B$ is $6$ and the distance from $P$ to $C$ is $5$. How to find the side of an equilateral triangle from this information?

You may be knowing this result for an equilateral triangle of side a and distance from P to vertices as x,y and z.
$$\frac{a^2+x^2-y^2}{2xa} = \frac{\sqrt{3}}{2}\frac{a^2+x^2-z^2}{2xa}+\frac{1}{2} \sqrt{1- \Big(\frac{a^2+x^2-z^2}{2xa}\Big)^2}$$
After putting values $(x,y,z)=(4,5,6)$ it becomes:
$$\frac{a^2+4^2-5^2}{8a} = \frac{\sqrt{3}}{2}\frac{a^2+4^2-6^2}{8a}+\frac{1}{2} \sqrt{1- \Big(\frac{a^2+4^2-6^2}{8a}\Big)^2}$$
Solving we get:
$$a=1/2\sqrt{154\pm30\sqrt{21}}\sim 8,2$$
