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2003年保加利亚国家数学奥林匹克(决赛)的第2题是:设H是锐角△ABC的高线CP上的任一点,直线AH、BH分别交BC、AC于点M、N.(1)证明:∠NPC=∠MPC.(2)设O是MN与CP的交点,一条通过O的任意的直线交四边形CNHM的边于D、E两点.证明:∠EPC=∠DPC.此题的证明可见文[1],给出的是一种三角证法,本文这里再给出该题的另一种解析证法,供赏析参考.证明如图,建立平面直角坐标系,不妨
The second question of the 2003 National Mathematical Olympiad of Bulgaria (final) is: Let H be any point on the high line CP of the acute angle △ ABC, and the lines AH and BH should be respectively BC and AC at points M and N. (1) Proof: ∠NPC = ∠MPC. (2) Let O be the intersection of MN and CP, an arbitrary straight line passing through O intersects the edge of CNHM at points D and E. Proof: ∠EPC = ∠DPC. Paper [1], given a triangular method, the article here is given another analysis of the problem of the card, for appreciation reference. Prove the map, the establishment of rectangular Cartesian coordinate system, may wish