In fact, we can take a point $S$ on $C R$ such that $R S = C P$. Then, from $\angle A C S = \angle Q P C = \angle B P R$, it can be deduced that $S C = R P$.

By the product of ratios in a complete quadrilateral (i.e., applying Menelaus' theorem to $\triangle A B Q$ and the transversal $R P C$), we know that $A C = B P$.

Furthermore, from $\triangle A C S \cong \triangle B R P$, we get $A D = B R$. This leads to $A S = A R = R S$, which means $\angle A R S = 60^{\circ}$, and thus $\angle B R C = 120^{\circ}$.
Test $3 P$ is a point inside $\triangle A B C$, line $A C$ and $B P$ intersect at $Q$, line $A B$ and $C P$ intersect at $R$, it is known that $A R=R B=C P, C Q=P Q$. Find $\angle B R C$.