8. Let $K$ and $L$ be the feet of perpendiculars from $P$ and $Q$ to $B C$ and $A C$ respectively. Let $M, N$ be points on $A B$ (ordered $A-N-M-B$) such that $R M N$ is a right isosceles triangle with $\angle R=90^{\circ}$. By sine theorem we have $\frac{B M}{B A}=\frac{B M}{B R} \cdot \frac{B R}{B A}=\frac{\sin 15^{\circ}}{\sin 45^{\circ}}$. Since $\frac{B K}{B C}=\frac{\sin 45^{\circ} \sin 30^{\circ}}{\cos 15^{\circ}}=\frac{\sin 15^{\circ}}{\sin 45^{\circ}}$, we deduce that $M K \| A C$ and $M K=A L$. Similarly, $N L \| B C$ and $N L=B K$. It follows that the vectors $\overrightarrow{R N}, \overrightarrow{N L}, \overrightarrow{L Q}$ are the images of $\overrightarrow{R M}, \overrightarrow{K P}, \overrightarrow{M K}$ respectively under a rotation of $90^{\circ}$, and consequently the same holds for their sums $\overrightarrow{R Q}$ and $\overrightarrow{R P}$. Therefore, $Q R=R P$ and $\angle Q R P=90^{\circ}$. Second solution. Let $A B S$ be the equilateral triangle constructed in the exterior of $\triangle A B C$. Obviously, the triangles $B P C, B R S, A R S, A Q C$ are similar. Let $f$ be the rotational homothety centered at $B$ that maps $P$ onto $C$, and let $g$ be the rotational homothety about $A$ that maps $C$ onto $Q$. The composition $h=g \circ f$ is also a rotational homothety; its angle is $\angle P B C+\angle C A Q=90^{\circ}$, and the coefficient is $\frac{B C}{B P} \cdot \frac{A Q}{A C}=1$. Moreover, $R$ is a fixed point of $h$ because $f(R)=S$ and $g(S)=R$. Hence $R$ is the center of $h$, and the statement follows from $h(P)=Q$. Remark. There are two more possible approaches: One includes using complex numbers and the other one is mere calculating of $R P, R Q, P Q$ by the cosine theorem. Second remark. The problem allows a generalization: Given that $\angle C B P=$ $\angle C A Q=\alpha, \angle B C P=\angle A C Q=\beta$, and $\angle R A B=\angle R B A=90^{\circ}-\alpha-\beta$, show that $R P=R Q$ and $\angle P R Q=2 \alpha$. !