Points $A,B,C$ and $D$ lie on a line, in that order, with $AB=CD$ and $BC=10$. Point $E$ is not on the line, and $BE=CE=13$. The perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$. Find $AB$.

1. Draw $EF\perp AD$; since $EB=EC=13$, $BF=FC=5$, so $EF=\sqrt{13^2-5^2}=12$ (theorem 18).
2. For right triangle $AEF$: $AE=\sqrt{(x+5)^2+12^2}$, where $x=AB$.
3. $DE=AE$ by symmetry, so the condition is $2\sqrt{(x+5)^2+12^2}+2x+10=2(13+13+10)=72$, i.e. $\sqrt{(x+5)^2+144}=31-x$. Squaring, $x^2+10x+169=961-62x+x^2$, so $72x=792$ and $x=\boxed{11}$.