Solution:

Let $D$ be the foot of the $A$-altitude, $E$ the midpoint of $AC$, and $F$ the foot of the $C$-internal angle bisector. Then by Ceva's Theorem, we have
$$\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA}=1,$$
and so
$$\frac{b}{a} \cdot \frac{c \cos B}{b \cos C}=1,$$
where we are using the shorthand $BC=a$, $CA=b$, $AB=c$. By cosine law, we know that
$$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}$$
and
$$\cos B=\frac{c^{2}+a^{2}-b^{2}}{2ca}.$$
Substituting this into the equation, we obtain
$$a\left(a^{2}+b^{2}-c^{2}\right)=b\left(c^{2}+a^{2}-b^{2}\right)$$
Solving for $c^{2}=AB^{2}$ in this equation then gives the final answer, which is $205$.