Completing the Square

First, we notice that $Eqn. 1$ is equivalent to $\sqrt{(x+d)^2+(y-c)^2} = \sqrt{(x-c)^2+(y-d)^2}$

Second, we notice that $Eqn. 2$ is equivalent to $((d-y)/(c-x))((c-y)/(-d-x))=-1$

Now, if we let $P(c, d)$ , $Q(-d, c)$ , and $R(x, y)$ , we can conclude, using the first two statements, that $PR = RQ$ and $PR \perp RQ$ . Thus, if $P$ and $Q$ were opposite vertices of a square, then $R$ would represent the other two vertices.

We notice that $x = y = 0$ would've been a solution, if not for the restriction that $x$ and $y$ are non-zero. Thus, $O(0, 0)$ is another vertex of the square.

Now, we construct $square$ $HJKL$ such that its sides are parallel to the axes and pass through $P$ , $O$ , $Q$ , and $R$ .

From the resulting figure, we conclude that $x = c - d$ and $y = c + d$ .

Thus, $x + y = 2c$ $\rightarrow$ $A=0$ , $B=0$ , $C=2$ , $D=0$ , $E=0$ $\rightarrow$ $[A+B+C+D+E]=\boxed{2}$

It can also be solved using algebra:

From the first condition we have that:

$y(c-d)=x(c+d)$ .

So we can let $x=k(c-d)$ and $y=k(c+d)$ where $k$ is some non-zero real number.

Expanding the second condition, we get:

$x^2+y^2=(c-d)x+(c+d)y$ .

Plugging in $x=k(c-d)$ and $y=k(c+d)$ , we get:

$((c-d)^2+(c+d)^2)k^2=((c-d)^2+(c+d)^2)k$

Since $c$ and $d$ are greater than $0$ , then $(c-d)^2+(c+d)^2$ must be greater than $0$ .

So we can divide by $(c-d)^2+(c+d)^2$ on both sides:

$k^2=k$ .

So $k=0$ or $k=1$ . Since $k$ must be non-zero, $k$ must be $1$ .