Let $A B C D$ be a quadrilateral with $\angle B A D=\angle A B C=90^{\circ}$, and suppose $A B=B C=1, A D=2$. The circumcircle of $A B C$ meets $\overline{A D}$ and $\overline{B D}$ at points $E$ and $F$, respectively. If lines $A F$ and $C D$ meet at $K$, compute $E K$.
Answer: $\sqrt{\frac{\sqrt{2}}{2}}$ Assign coordinates such that $B$ is the origin, $A$ is $(0,1)$, and $C$ is $(1,0)$. Clearly, $E$ is the point $(1,1)$. Since the circumcenter of $\mathrm{ABC}$ is $\left(\frac{1}{2}, \frac{1}{2}\right)$, the equation of the circumcircle of $A B C$ is $\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{2}$. Since line $B D$ is given by $x=2 y$, we find that $F$ is at $\left(\frac{6}{5}, \frac{3}{5}\right)$. The intersection of $A F$ with $C D$ is therefore at $\left(\frac{3}{2}, \frac{1}{2}\right)$, so $K$ is the midpoint of $C D$. As a result, $E K=\frac{\sqrt{2}}{2}$. This is in fact a special case of APMO 2013, Problem 5, when the quadrilateral is a square.