Answer: 463

Solution:

A cevian $AD$ splits the opposite side in the ratio $$\begin{aligned} \frac{BD}{DC} &= \frac{[ABD]}{[ACD]} \\ &= \small \frac{\frac{1}{2} AB \cdot AD \sin\angle BAD}{\frac{1}{2} AC \cdot AD \sin\angle CAD} \\ &= \frac{AB \sin\angle BAD}{AC \sin\angle CAD}, \end{aligned}$$ and similarly $\frac{BE}{EC} = \frac{AB \sin\angle BAE}{AC \sin\angle CAE}.$

Since $\angle BAE = \angle CAD,$ we also have $\angle BAD = \angle CAE$ (each is that common angle plus $\angle EAD$), so multiplying the two ratios cancels all the sines: $\frac{BD}{DC} \cdot \frac{BE}{EC} = \frac{AB^2}{AC^2}.$ With $BD = 9,$ $DC = 6,$ $AB = 13,$ and $AC = 14,$ this gives $$\frac{BE}{EC} = \frac{13^2}{14^2} \cdot \frac{6}{9} = \frac{169}{294}.$$

Hence $BE = 15 \cdot \frac{169}{169 + 294} = \frac{2535}{463}.$ Since $463$ is prime and does not divide $2535 = 3 \cdot 5 \cdot 13^2,$ the fraction is in lowest terms, and $q = 463.$