Let $\overline{AB}$ be a diameter in a circle of radius $5$. Let $\overline{CD}$ be a chord meeting $\overline{AB}$ at $E$ with $BE=2$ and $\angle AEC=45^\circ$. What is $CE^2+DE^2$?

1. As in the problem, let $O$ be the center, $F$ the foot of the perpendicular from $O$ to $CD$, and $OE=t$. Since $\angle AEC=45^\circ$, $OF=EF=\frac t{\sqrt2}$. Then $CE^2+DE^2=(CF-EF)^2+(CF+EF)^2=2CF^2+2EF^2=2(R^2-OF^2)+2EF^2=2R^2$, wherever $E$ lies on the diameter.
2. With $R=5$: $CE^2+DE^2=2(25)=\boxed{50}$.