In triangle $ABC$, $AC = 13$, $BC = 14$,  and $AB=15$. Points $M$ and $D$ lie on $AC$ with $AM=MC$ and $\angle ABD = \angle DBC$. Points $N$ and $E$ lie on $AB$ with $AN=NB$ and $\angle ACE = \angle ECB$. Let $P$ be the point, other than $A$, of intersection of the circumcircles of $\triangle AMN$ and $\triangle ADE$. Ray $AP$ meets $BC$ at $Q$. The ratio $\frac{BQ}{CQ}$ can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m-n$.
Define the function $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by \[f(X)=\text{Pow}_{(AMN)}(X)-\text{Pow}_{(ADE)}(X)\] for points $X$ in the plane. Then $f$ is linear, so $\frac{BQ}{CQ}=\frac{f(B)-f(Q)}{f(Q)-f(C)}$. But $f(Q)=0$ since $Q$ lies on the radical axis of $(AMN)$, $(ADE)$ thus \[\frac{BQ}{CQ}=-\frac{f(B)}{f(C)}=-\frac{BN\cdot BA-BE\cdot BA}{CM\cdot CA-CD\cdot CA}\] Let $AC=b$, $BC=a$ and $AB=c$. Note that $BN=\tfrac{c}{2}$ and $CM=\tfrac{b}{2}$ because they are midpoints, while $BE=\frac{ac}{a+b}$ and $CD=\frac{ab}{a+c}$ by Angle Bisector Theorem. Thus we can rewrite this expression as \begin{align*}&-\frac{\tfrac{c^{2}}{2}-\tfrac{ac^{2}}{a+b}}{\tfrac{b^{2}}{2}-\tfrac{ab^{2}}{a+c}} \\ =&-\left(\frac{c^{2}}{b^{2}}\right)\left(\frac{\tfrac{1}{2}-\tfrac{a}{a+b}}{\tfrac{1}{2}-\tfrac{a}{a+c}}\right) \\ =&\left(\frac{c^{2}}{b^{2}}\right)\left(\frac{a+c}{a+b}\right) \\ =&\left(\frac{225}{169}\right)\left(\frac{29}{27}\right) \\ =&~\frac{725}{507}\end{align*} so $m-n=\boxed{218}$.