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

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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如图,△ABC中,AB=AC,AD⊥BC于点D,BE⊥AC于点E,AD与BE相交于

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
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

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
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