The angle condition implies that $M,N$ are isogonal conjugates. We now work in the complex plane. Then $\frac{AM\cdot AN}{AB\cdot AC}=\frac{(a-m)(a-n)}{(a-b)(a-c)}$ (since this is guaranteed to be real by the angle condition). This yields \begin{align*}\frac{AM\cdot AN}{AB\cdot AC}+\frac{BM\cdot BN}{BA\cdot BC}+\frac{CM\cdot CN}{CA\cdot CB}&=\frac{(a-m)(a-n)}{(a-b)(a-c)}+\frac{(b-m)(b-n)}{(b-c)(b-a)}+\frac{(c-m)(c-n)}{(c-a)(c-b)}\\&=\frac{-\sum\limits_\text{cyc}a^2b-a^2c+(m+n)(ac-ab)+mn(b-c)}{(a-b)(b-c)(c-a)}\\&=\frac{-\sum\limits_\text{cyc}a^2b-a^2c}{(a-b)(b-c)(c-a)}=1,\end{align*}as desired.

Davsch wrote: since this is guaranteed to be real by the angle condition Could you please provide more detailed proof

If you take reflections of $N$ over the sides of the triangle you get a hexagon having twice the area of the triangle and 6 triangles , 2 of them having angle $\hat A$ between $AM, AN$, 2 having angle $\hat B$ between $BM, BN$ and other 2 having angle $\hat C$ between $CM, CN$, and so you can get a nice solution by summing up areas ratios. That proof I saw it in a book by V. Vornicu. Best regards, sunken rock

See ISL 1998/G4.