如图,设P是△ABC内任一点,AD,BE,CF是过点P且分别交边BC,CA,AB于D,E,F. 求证:PD/AD+PE/BE+PF/CF=1.

问题描述:

如图,设P是△ABC内任一点,AD,BE,CF是过点P且分别交边BC,CA,AB于D,E,F.
求证:

PD
AD
+
PE
BE
+
PF
CF
=1.

证明:∵S△BDP:S△ABD=DP:AD,S△CDP:S△ACD=DP:AD,∴(S△BDP+S△CDP):(S△ABD+S△ACD)=DP:AD,∴S△BCP:S△ABC=DP:AD①,同理S△ABP:S△ABC=PF:CF②,S△ACP:S△ABC=PE:BE③,①+②+③,得(S△B...