Question

几何题：P为正△ABC内任意一点，P到AB,AC,BC距离分别为PE,PF,PD,连接AP,BP,CP

P为正△ABC内任意一点，P到AB,AC,BC距离分别为PE,PF,PD,连接AP,BP,CP求证S△APF+S△CPD+S△BPE=1/2S△ABC

Answer 1

如图所示，过点P作MN//BC，分别交AB,AC于M,N

过点P作XY//AC，分别交BA,BC于X,Y

过点P作UV//AB，分别交CB,CA于U,V

则易知△PVN，△PMX，△PUY都是等边△

∵PD,PE,PF分别⊥UY,MX,VN

∴VF=FN，ME=EX，UD=DY

∴S△PFV=S△PFN，S△PEM=S△PEX，S△PDU=S△PDY

又PVAX，PMBU，PYCN都是平行四边形

∴S△PAV=S△PAX，S△PBM=S△PBU，S△PCY=S△PCN

∴S△APF+S△CPD+S△BPE

=(S△PAV+S△PFV)+(S△PCY+S△PDY)+(S△PBM+S△PEM)

=(S△PAX+S△PFN)+(S△PCN+S△PDU)+(S△PBU+S△PEX)

=(S△PAX+S△PEX)+(S△PCN+S△PFN)+(S△PBU+S△PDU)

=S△PAE+S△PCF+S△PBD

而(S△APF+S△CPD+S△BPE)+(S△PAE+S△PCF+S△PBD)=S△ABC

∴S△APF+S△CPD+S△BPE=(1/2)S△ABC