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_数学解答

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

人气：456 ℃　时间：2020-02-01 08:39:29

解答

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