in triangle ABC The point P lies on the side BC so that AB+BP=AC+CP ,and the point Q lies on the side AC so that:BA+AQ=BC+CQ and the point R lies on the side AB so that:CB+BR=CA+AR .How to prove that the lines AP,BQ,CR are meet
in triangle ABC The point P lies on the side BC so that AB+BP=AC+CP ,and the point Q lies on the side AC so that:BA+AQ=BC+CQ and the point R lies on the side AB so that:CB+BR=CA+AR .How to prove that the lines AP,BQ,CR are meet
Suppose AR = a, RB = b, BP = c, PC = d, CQ = e, QA = f. From the conditions of the problem, we know that
(1) $\displaystyle a+b+c = d+e+f$
(2) $\displaystyle b+c+d = e+f+a$
(3) $\displaystyle c+d+e = f+a+b$
Subtracting (2) from (1), we obtain $\displaystyle a-d = d-a \Rightarrow a = d$. Subtracting (3) from (2), $\displaystyle b-e = e-b \Rightarrow b = e$. Plugging this back into (1), we obtain $\displaystyle c = f$.
Therefore $\displaystyle \frac{a}{b}*\frac{c}{d}*\frac{e}{f} = \frac{a}{b}*\frac{c}{a}*\frac{b}{c} = 1$. The segments AP, BQ, CR must intersect at a point, due to Ceva's theorem.