ABC, the non-degenerate triangle

Let $R$ be the radius of the circumcircle of triangle $ABC$ and $BC=a$ , $AC=b$ , $AB=c$ , then by the extended sine rule, we have $\sin \angle A = \frac{a}{2R}$ , $\sin \angle B = \frac{b}{2R}$ and $\sin C = \frac{c}{2R}$ . Substituting this into the given equation, we have

$\begin{aligned} 2\sin \angle B \cdot \cos \angle C + \sin \angle C &= \sin A + \sin B \\ 2\cdot \frac{b}{2R} \cdot \cos \angle C + \frac{c}{2R} &= \frac{a}{2R} + \frac{b}{2R} \\ \cos \angle C &= \frac{a + b-c}{2b} \\ \end{aligned}$

From the cosine rule, we have $\cos \angle C = \frac{a^2+b^2-c^2}{2ab}$ .

Equating the two we have

$\begin{aligned} \frac{a + b-c}{2b} &= \frac{a^2+b^2-c^2}{2ab} \\ a^2 + ab - ac &= a^2 + b^2 - c^2 \\ 0 &= b^2-c^2 - ab + ac \\ &= (b+c-a)(b-c) \\ \end{aligned}$

By the triangle inequality, we have $b + c > a$ for non-degenerate triangles, thus $b=c$ is the only possible solution. Hence $100\frac{AC}{AB} = 100\frac{b}{c} = 100$ .