题目内容
如图,锐角△ABC中,高BE、CF交于点H.
(1)若∠BAC=70°,求∠BHC的度数;
(2)直接给出四条线段AF、HE、AC、CH之间的数量关系;
(3)若AD平分∠BAC交BC于D,AD、CF交于点K,HG平分∠BHC交BC于G.求证:HG∥AD.
(1)若∠BAC=70°,求∠BHC的度数;
(2)直接给出四条线段AF、HE、AC、CH之间的数量关系;
(3)若AD平分∠BAC交BC于D,AD、CF交于点K,HG平分∠BHC交BC于G.求证:HG∥AD.
(1)在四边形AFHE中,∠AFH=∠AEH=90°,∠BAC=70°,
∴∠BHC=∠FHE=360°-(∠AFH+∠AEH+∠BAC),
=360°-250°,
=110°;
或∠BHC=∠HEC+∠ACH=90°+(90°-∠FAC)=180°-70°=110°;
(2)∵∠ACF=∠HCE,∠AFC=∠HEC=90°,
∴△ACF∽△HCE,
∴
=
,或AF•HC=HE•AC;
(3)由(1)得∠BHC=360°-(90°+90°+∠BAC),
=180°-∠BAC,
又∵HG平分∠BHC,
∴∠GHC=
∠BHC=90°-
∠BAC,
∠DKH=∠AKF=90°-∠FAK=90°-
∠BAC,
∴∠GHC=∠DKH,
∴HG∥AD.
∴∠BHC=∠FHE=360°-(∠AFH+∠AEH+∠BAC),
=360°-250°,
=110°;
或∠BHC=∠HEC+∠ACH=90°+(90°-∠FAC)=180°-70°=110°;
(2)∵∠ACF=∠HCE,∠AFC=∠HEC=90°,
∴△ACF∽△HCE,
∴
AF |
HE |
AC |
HC |
(3)由(1)得∠BHC=360°-(90°+90°+∠BAC),
=180°-∠BAC,
又∵HG平分∠BHC,
∴∠GHC=
1 |
2 |
1 |
2 |
∠DKH=∠AKF=90°-∠FAK=90°-
1 |
2 |
∴∠GHC=∠DKH,
∴HG∥AD.
练习册系列答案
相关题目