For example, (2003 Japan Mathematical Olympiad) $P$ is a point inside $\triangle A B C$, the line $A C$ and $B P$ intersect at $Q$, the line $A B$ and $C P$ intersect at $R$. Given that $A R=R B=C P, O Q=P Q$. Try to find the size of $\angle B R C$.

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The translation is provided as requested, maintaining the original text's line breaks and format.
Analysis: Combining the diagram, guess first and then prove.
Solution: Let $S$ be a point on line segment $CR$ such that $RS = CP$.
Since $CQ = PQ$,
we have $\angle ACS = \angle QPC = \angle BPR$.
Since $RS = CP$,
we have $SC = CR - RS = CR - CP = RP$.
Considering $\triangle ABQ$ and line $CR$, by Menelaus' theorem, we have
$$
\frac{AC}{CQ} \cdot \frac{QP}{PB} \cdot \frac{BR}{RA} = 1
$$

Since $AR = RB$ and $CQ = PQ$,
we have $AC = BP$.
Therefore, $\triangle ACS \cong \triangle BPR$.
Thus, $AS = BR$.
Since $RS = CP$ and $AR = RB = CP$,
we have $AS = AR = RS$.
Therefore, $\angle ARS = 60^{\circ}$.
Thus, $\angle BRC = 120^{\circ}$.