 Solution

Let $r=\arcsin (\sin \theta /n)$ be the angle inside the prism relative to the inward normal of face A. For the entry position shown, below the midpoint of A, and $0<\theta <90^{\circ }$, the transmitted ray first reaches face B. Its incidence angle there is $i_B=|\alpha -r|$. If $n\sin i_B<1$, a transmitted ray leaves B according to Snell's law. If $n\sin i_B>1$, it undergoes total internal reflection at B and then reaches C. At the equality, transmission is grazing.

On the reflected route through this equilateral prism, the incidence angle at C is $r$, so $n\sin r=\sin \theta <1$ permits emergence from C. The one-reflection path through an equilateral prism is drawn alongside the direct-transmission route.

Real interfaces also produce partial Fresnel reflection below the critical angle; these additional weaker rays are omitted from the principal-path sketch. Which face a ray first reaches also depends on its entrance position, so the route statement uses the entrance geometry illustrated in the source.