Geometry Meets Number Theory

Without loss of generality, put $A$ at the origin and $B$ at $(b,0).$ The condition given for $C,D,E$ is that they all satisfy the equation $x^2+y^2 = 2((x-b)^2+y^2),$ which becomes, after some manipulation, $(x-2b)^2+y^2 = 2b^2.$ This is a circle centered at $(2b,0)$ with radius $b \sqrt{2}.$ Since $CDE$ is a right triangle whose vertices lie on the circle, ${\overline{CE}}$ is a diameter, of length $2b\sqrt{2},$ and we get $CD^2 + DE^2 = 8b^2.$ If $CD,DE,b$ are all integers (note $b=AB$ ), then looking mod $8$ shows that $CD,DE$ are both even, say $CD = 2x, DE = 2y,$ and the equation becomes $x^2+y^2=2b^2$ where $x,y$ are distinct positive integers (and neither equals $b/2$ ).

The set of solutions to this equation can be parameterized, but as we are only looking for the minimal solution, we can just search over small values of $b.$ The smallest value of $b$ where a solution exists with positive, distinct $x,y$ is $b=5, x=1, y =7,$ which leads to $AB = 5, CD = 2, DE = 14,$ and a sum of $\fbox{21}.$ There are no other such solutions for $b \le 10,$ so it's clear that this gives the minimal sum.