Square $ABCD$ has side length $2$. A semicircle with diameter $\overline{AB}$ is constructed inside the square, and the tangent to the semicircle from $C$ intersects side $\overline{AD}$ at $E$. What is the length of $\overline{CE}$? $\mathrm{(A) \ } \frac{2+\sqrt{5}}{2} \qquad$ $ \mathrm{(B) \ } \sqrt{5} \qquad$ $ \mathrm{(C) \ } \sqrt{6} \qquad$ $ \mathrm{(D) \ } \frac{5}{2} \qquad$ $ \mathrm{(E) \ } 5-\sqrt{5}$
Solution
Assume $EA=EF=x$, we know $CF=CB=2$.
By Pythagorean theorem with triangle $CED$, we have
$(2-x)^2+2^2=(2+x)^2$, so $x=\frac12$, and $CE=\frac52$.
Rectangle $ABCD$ has $AB=6$ and $BC=8$. A semicircle with diameter $\overline{AB}$ is constructed inside the rectangle, and the tangent to the semicircle from $C$ (other than $\overline{CB}$) intersects side $\overline{AD}$ at $E$. What is the length of $\overline{CE}$?
$\mathrm{(A) \ } \frac{2+\sqrt{5}}{2} \qquad$ $ \mathrm{(B) \ } \sqrt{5} \qquad$ $ \mathrm{(C) \ } \sqrt{6} \qquad$ $ \mathrm{(D) \ } \frac{5}{2} \qquad$ $ \mathrm{(E) \ } 5-\sqrt{5}$ 
