(1)图1做BF⊥EC于F 图2做BH⊥EC于H
①结论:BD=CE,BD⊥CE;
②结论:BD=CE,BD⊥CE
理由如下:∵∠BAC=∠DAE=90°
∴∠BAD-∠DAC=∠DAE-∠DAC,即∠BAD=∠CAE
在△ABD与△ACE中,
∵AB=AC∠BAD=∠CAEAD=AE
∴△ABD≌△ACE
∴BD=CE
延长BD交AC于F,交CE于H.
在△ABF与△HCF中,
∵∠ABF=∠HCF,∠AFB=∠HFC
∴∠CHF=∠BAF=90°
∴BD⊥CE
(2)结论:乙.AB:AC=AD:AE,∠BAC=∠DAE=90°