如图△abe中ab=ae,ad=ac,角bad=角eac,bc,ed交与点o

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如图△abe中ab=ae,ad=ac,角bad=角eac,bc,ed交与点o如图△abe中ab=ae,ad=ac,角bad=角eac,bc,ed交与点o如图△abe中ab=ae,ad=ac,角bad=

如图△abe中ab=ae,ad=ac,角bad=角eac,bc,ed交与点o
如图△abe中ab=ae,ad=ac,角bad=角eac,bc,ed交与点o

如图△abe中ab=ae,ad=ac,角bad=角eac,bc,ed交与点o
(1)证明:因为∠BAD=∠EAC, 则∠BAC=∠EAD, 又因为AB=AE,AC=AD, 所以△ABC≌△AED. (2)因为△ABC≌△AED, 所以∠ABC=∠AED, 又因为AB=AE, 所以∠ABE=∠AEB, 所以∠ABE-∠ABC=∠AEB-∠AEB, 即∠CBE=∠DEB.