解答:解:(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⊥C
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
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追答![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b3119313b07eca8042333184922397dda144839a?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
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我谢谢了