Answer: Yes; it is an equilateral triangle (each side 4).
\[AB^2=(2\sqrt2)^2+(2\sqrt2)^2=8+8=16\]
\[BC^2=(\sqrt6-\sqrt2)^2+(\sqrt6+\sqrt2)^2\]
\[=2(6+2)=16\]
\[AC^2=(\sqrt6+\sqrt2)^2+(\sqrt6-\sqrt2)^2=16\]
(using \((a-b)^2+(a+b)^2=2(a^2+b^2)\)). So AB = BC = AC = 4. Equal sides cannot satisfy "sum of two = third", so a triangle is formed, and it is equilateral.