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)
(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.