G为锐角三角形ABC重心P为内部任意一点PG交BC延长线于A1交CA于C1交AB于B1求证A1P÷A1G+B1P÷B1G+C1P÷C1G=3

Question

G为锐角三角形ABC重心 P为内部任意一点PG交BC延长线于A1 交CA于C1 交AB于B1
求证A1P÷A1G+B1P÷B1G+C1P÷C1G=3

Answer 1

连接BG,GC,PB,PC,分别过G,P作GG1⊥BC于G1,作PP1⊥BC于P1,则PP1‖GG1,PP1/GG1=A1P/A1G.
S(△PBC)/S(△GBC)=PP1/GG1=A1P/A1G.
同理,S(△PCA)/S(△GCA)=C1P/C1G.
S(△PAB)/S(△GAB)=B1P/B1G.
G为重心,S(△GAB)=S(△GBC)=S(△GCA)=S(△ABC)/3.
所以,A1P/A1G+B1P/B1G+C1P/C1G=
=3S(△PBC)/S(△ABC)+3S(△PAB)/S(△ABC)+3S(△PCA)/S(△ABC)=3.