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)

(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.