H为△ABC的垂心,AH⊥BC于D,BH⊥AC于E,CH⊥AB于F,求证：DH/DA+EH/EB+FH/FC＝1

题目

H为△ABC的垂心,AH⊥BC于D,BH⊥AC于E,CH⊥AB于F,求证：DH/DA+EH/EB+FH/FC＝1

答案

S△HBC/S△ABC=½BC·DH/(½BC·DA)=DH/DA,则：
DH/DA+EH/EB+FH/FC
=S△HBC/S△ABC+S△AHC/S△ABC+S△AHB/S△ABC
=(S△HBC+S△AHC+S△AHB)/S△ABC
=S△ABC/S△ABC
=1