Extend the cevians $AX$, $BX$, $CX$; let their intersection points with the sides $BC$, $AC$, $BA$ be $A_1$, $B_1$, $C_1$ respectively. We have
$$\angle AXB_1 = 180^\circ - \angle BXA = 180^\circ - (\angle B + \angle C) = \angle A = \alpha.$$
Similarly, $\angle BXC_1 = \angle B = \beta$, $\angle CXA_1 = \angle C = \gamma$. Let $\angle CAA_1 = \varphi$. Then $\angle ABB_1 = \angle ABX = 180^\circ - (\alpha - \varphi) - (\alpha + \beta) = \varphi$ and, similarly, $\angle BCC_1 = \varphi$. That is, $X$ is the Brakar point of the triangle $ABC$.

Now, using the law of sines for the triangles $ABC$, $AXB$, $AXC$, we have respectively
$$\frac{BC}{AC} = \frac{\sin \alpha}{\sin \gamma}, \quad \frac{AX}{AB} = \frac{\sin \varphi}{\sin \angle BXA} = \frac{\sin \varphi}{\sin \alpha},$$
$$\frac{AC}{CX} = \frac{\sin \angle CXA}{\sin \varphi} = \frac{\sin \gamma}{\sin \varphi}.$$
Multiplying all three equalities we obtain
$$\frac{BC}{AB} \cdot \frac{AX}{AB} \cdot \frac{AC}{CX} \text{ which is equivalent to } \frac{BC \cdot AX}{AB} = \frac{AB \cdot CX}{AC}.$$