In the coordinate plane, $A(0,3)$, $B(6,0)$, $D(1,0)$, and $C(5,2)$ are given. Segment $AB$ meets segment $CD$ at $E$. Find the length of $AE$.

Line $DC$ extended meets the horizontal line through $A$ (i.e. $y=3$) at $F$. Since $CD$ has slope $\frac{2-0}{5-1}=\frac12$, from $C(5,2)$ we get $F=(7,3)$.
Then $AF=7$ (horizontal distance from $A(0,3)$), and $DB=6-1=5$, so $AF:DB=7:5$.
Since $AF\parallel DB$, $\triangle AEF\sim \triangle BED$, so $AE:BE=AF:DB=7:5$, giving $AE:AB=7:12$.
$AB=\sqrt{6^2+3^2}=3\sqrt5$, so $AE=\frac7{12}(3\sqrt5)=\boxed{\frac{7\sqrt5}4}$.