Suppose AR = a, RB = b, BP = c, PC = d, CQ = e, QA = f. From the conditions of the problem, we know that

(1)

(2)

(3)

Subtracting (2) from (1), we obtain . Subtracting (3) from (2), . Plugging this back into (1), we obtain .

Therefore . The segments AP, BQ, CR must intersect at a point, due to Ceva's theorem.