10. $A$ is the center of a semicircle, with radius $A D$ lying on the base. $B$ lies on the base between $A$ and $D$, and $E$ is on the circular portion of the semicircle such that $E B A$ is a right angle. Extend $E A$ through $A$ to $C$, and put $F$ on line $C D$ such that $E B F$ is a line. Now $E A=1, A C=\sqrt{2}, B F=\frac{2-\sqrt{2}}{4}, C F=\frac{2 \sqrt{5}+\sqrt{10}}{4}$, and $D F=\frac{2 \sqrt{5}-\sqrt{10}}{4}$. Find $D E$.
Solution: Let $\theta=\angle A E D$ and $x=D E$. By the law of cosines on triangle $A D E$, we have $1=1+x^{2}-2 x \cos \theta \Rightarrow 2 x \cos \theta=x^{2}$. Then by the law of cosines on triangle $C D E$ (note that $C D=\sqrt{5}$ ), we have $5=(1+\sqrt{2})^{2}+x^{2}-2(1+\sqrt{2}) x \cos \theta=(1+\sqrt{2})^{2}+x^{2}-(1+\sqrt{2}) x^{2}$. Solving the quadratic equation gives $x=\sqrt{2-\sqrt{2}}$.