Question

如图,△ABC中,AB=AC,AD⊥BC于点D,BE⊥AC于点E,AD与BE相交于点H,AE=BE,试说明AH＝2BD

Answer 1

证明:延长BE使HE=EF,连接AF

BE⊥AC,HE=EF,易知△AHF是等腰三角形

∠HAE=∠EAF,AH=AF

AD⊥BC,BE⊥AC

∠HDB=∠AEH=90°

因∠HBD+∠BHD=∠HAE+∠AHE=90°,

∠BHD=∠HAE

所以∠HBD=∠AHE又∠HAE=∠EAF

∠HBD=∠EAF,BE=AE,∠BEC=∠AEF=90°

RT△BEC≌RT△AEF(ASA)

BC=AF

AH=AF

BC=AH,BC=2BD

AH=2BD