In rectangle $ABCD$, $AB=28$ and $BC=15$. Let $E$ be a point on $\overline{CD}$ such that $BE=17$. What is $AE$?

1. In the right triangle $BCE$ (right angle at $C$), $CE=\sqrt{BE^2-BC^2}=\sqrt{17^2-15^2}=\sqrt{64}=8$.
2. Since $CD=AB=28$, we get $DE=CD-CE=28-8=20$, so $E$ lies strictly between $D$ and $C$.
3. In the right triangle $ADE$ (right angle at $D$), $AD=BC=15$, so $AE=\sqrt{AD^2+DE^2}=\sqrt{15^2+20^2}=\sqrt{625}=\boxed{25}$.
4. (The law of cosines gives the same value: $\cos\angle ABE=\frac{CE}{BE}=\frac8{17}$, so $AE^2=28^2+17^2-2\cdot28\cdot17\cdot\frac8{17}=625$.)