摘要：23． 已知:△ABC是任意三角形． ⑴如图1所示.点M.P.N分别是边AB.BC.CA的中点．求证:∠MPN=∠A． ⑵如图2所示.点M.N分别在边AB.AC上.且..点P1.P2是边BC的三等分点.你认为∠MP1N+∠MP2N=∠A是否正确?请说明你的理由． ⑶如图3所示.点M.N分别在边AB.AC上.且..点P1.P2.--.P2009是边BC的2010等分点.则∠MP1N+∠MP2N+--+∠MP2009N= ． (请直接将该小问的答案写在横线上．) 答案:23. ⑴证明:∵点M.P.N分别是AB.BC.CA的中点. ∴线段MP.PN是△ABC的中位线. ∴MP∥AN.PN∥AM.················· 1分 ∴四边形AMPN是平行四边形.····· 2分 ∴∠MPN=∠A. ·························· 3分 ⑵∠MP1N+∠MP2N=∠A正确. ················· 4分 如图所示.连接MN. ························· 5分 ∵.∠A=∠A. ∴△AMN∽△ABC. ∴∠AMN=∠B.. ∴MN∥BC.MN=BC. ····················· 6分 ∵点P1.P2是边BC的三等分点. ∴MN与BP1平行且相等.MN与P1P2平行且相等.MN与P2C平行且相等. ∴四边形MBP1N.MP1P2N.MP2CN都是平行四边形. ∴MB∥NP1.MP1∥NP2.MP2∥AC. ································································· 7分 ∴∠MP1N=∠1.∠MP2N=∠2.∠BMP2=∠A. ∴∠MP1N+∠MP2N=∠1+∠2=∠BMP2=∠A. ······························································· 8分 ⑶∠A. ················································ 9分

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