Let the intersection of $BD$ and $AE$ be $G$, the centroid of $\triangle ABC$. Let the midpoints of $BD$ and $AE$ be $M$ and $N$. Then $AG=\frac{2}{3}AE$ and $AN=\frac{1}{2}AE$, so $GN=\frac{1}{6}AE=\frac{1}{4}AG$. Similarly, $GM=\frac{1}{4}BG$. Then we have similar triangles $GNM$ and $GAB$ with common ratio $\frac{1}{4}$, so $AB=4NM=4\cdot4.5=\boxed{18}$

With vectors, call $ X $ the midpoint of $ AE $ and $ Y $ the midpoint of $ BD $ , we have that $ \vec{X}=\frac{\vec{C}+\vec{B}}{4} $ and in the same way $ \vec{Y}=\frac{\vec{A}+\vec{C}}{4} $ , hence $ \vec{YX}=\vec{X}-\vec{Y}=\frac{\vec{C}+\vec{B}}{4}-\frac{\vec{A}+\vec{C}}{4}=\frac{\vec{B}-\vec{A}}{4}=\frac{\vec{AB}}{4} $ , so $ AB=4XY=18 $ GJOGH