A paper triangle $ABC$ has a right angle at $C$, with $AC=12$ inches and $BC=5$ inches (so $AB=13$ inches). The triangle is folded so that point $A$ falls exactly on point $B$. What is the length in inches of the crease?

The crease $DE$ is the perpendicular bisector of $AB$, so $D$ is the midpoint of $AB$ and $AD=\frac{13}2$. Triangles $AED$ and $ABC$ have right angles at $D$ and $C$ and share $\angle A$, so $\triangle AED\sim \triangle ABC$.
Then $DE:BC=AD:AC$, so $DE=\frac{5\cdot\frac{13}2}{12}=\boxed{\frac{65}{24}}$.