smc

We already know two vertices of the square: $(A+B)/2=(2,5)$ and $(B+C)/2=(3,2)$. There are only two possibilities for the other vertices of the square: either they are $(6,3)$ and $(5,6)$, or they are $(0,1)$ and $(-1,4)$. The second case would give us $D$ outside the first quadrant, hence the first case is the correct one. As $(6,3)$ is the midpoint of $CD$, we can compute $D=(7,3)$, and $7+3=\boxed{10}$.