Triangles on a Square

Constructing two external triangles congruent to $\triangle{AFB}$ and $\triangle{CED}$ with bases $\overline{AD}$ and $\overline{BC}$ makes this problem very simple. Consider one of the four congruent triangles and notice that it must be a right triangle because its sides represent a Pythagorean triple. Therefore, all four triangles are right triangles, and ultimately there is a square circumscribing square $ABCD$ with diagonal $\overline{EF}$ . By adding side lengths, we know that the side length of the larger square is $17$ . It follows that $\overline{EF} = 17\sqrt{2}$ which shows that $EF^{2} = \boxed{578}$