Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$ $\textbf{(A)}\ 96 \qquad$ $\textbf{(B)}\ 98 \qquad$ $\textbf{(C)}\ 44\sqrt5 \qquad$ $\textbf{(D)}\ 70\sqrt2 \qquad$ $\textbf{(E)}\ 100$
Solution
Draw $OF\bot CD$ at $F$, $BE=2\sqrt5, AE=10\sqrt2-2\sqrt5$.
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 $CE^2+DE^2$?
