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The above figure was done using grapher on mac osx.
Let $l$ be the length (i.e. the sides $AD$ and $BC$) and $b$ be the breadth (i.e. the sides $AB$ and $CD$). Once you get the diagonal vertices together, i.e. when $D$ coincides with $B$, the length $EB$ is the same as the length $ED = l-a$.
Hence, for the right triangle, we have that $a^2 + b^2 = (l-a)^2\\ b^2 = l^2 - 2al\\ a = \frac{l^2 - b^2}{2l}$ The length of $BF$ is $l-a$ and is given by $l-a = l - \frac{l^2 - b^2}{2l} = \frac{2l^2 - l^2 + b^2}{2l} = \frac{l^2 + b^2}{2l}$ The distance between the two points is $EF^2 = r^2 = b^2 + (l-2a)^2 = b^2 + \left(l - \frac{l^2 - b^2}{l} \right)^2 = b^2 + \frac{b^4}{l^2}$ This is so since the vertical distance between $E$ and $F$ is $l-2a$.
You are given that $r = l$ and hence $l^2 = b^2 + \frac{b^4}{l^2}$If we let $\frac{l}{b} = x,$ then we get that $x^2 = 1 + \frac1{x^2}\\ x^4 = x^2 + 1$ which gives us that $x = \sqrt{\frac{1}{2} (1+\sqrt{5})} = \sqrt{\phi}$ where $\phi$ is the golden ratio.