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

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- Jul 12th 2012, 03:12 AMMhmh96Proof
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

- Jul 12th 2012, 09:52 AMrichard1234Re: Proof
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.