Question

如图,AD、BE是△ABC的高,AD和EB的延长线相交于H,连结HC,且AH=BC,说明：∠CAB=∠HCA

Answer 1

∵AD⊥CD, HE⊥AC

∴∠HEC=∠HDC,

又∵∠CBE=∠HBD

∴∠BCE=∠BHD

∵AH=BC, ∠BCE=∠BHD, ∠CBE=∠HBD

∴△CBE≌△HAE

∴HE=AE, BE=AE

∴∠CAB=∠ABE=45°

∠HCA=∠CHE=45°

∴∠CAB=∠HCA