In trapezoid $ABCD$, $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$, with $AB+CD=BC$, $AB<CD$, and $AD=10$. What is $AB\cdot CD$?

1. Draw $BE\bot CD$.
2. $BE^2+CE^2=BC^2$, $10^2+(CD-AB)^2=(CD+AB)^2$, $AB\cdot CD=25$.
3. Answer is $25$.