题目:如图1,已知P为锐角△ABC内一点,过P分别作BC,AC,AB的垂线,垂足分别为D,E,F,BM为∠ABC的平分线,MP的延长线交AB于点N.如果PD=PE+PF,求证:CN是∠ACB的平分线.证法1:过N作NQ⊥AC于Q,NH⊥BC于H,过M作ML⊥AB于L,MR⊥BC于R,连NR交PD于G.因为BM平分∠ABC,所以ML=MR.又PF∥ML,PG∥ Title: As shown in Figure 1, it is known that P is a point within an acute angle △ ABC, P is the vertical line of BC, AC and AB respectively, and the vertical lines are D, E and F respectively. BM is the bisector of ∠ABC and MP extends Line intersection AB at point N. If PD = PE + PF, verify that: CN is the bisector of ∠ACB. Law 1: N over N for N, Q on ⊥, NH ⊥ BC on H, M over M for ML ⊥ AB L, MR⊥BC in R, even with NR PD in G. Because BM bisects ∠ABC, so ML = MR. Also PF∥ML, PG ∥