真么写已知如图ab=ac ad=ae ab dc相交于点m ac be相交于点n,角bab=角eac 求角am=an已知如图ab=ac ad=ae ab dc相交于点m ac be相交于点n,角bab=角eac求角am=an

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真么写已知如图ab=ac ad=ae ab dc相交于点m ac be相交于点n,角bab=角eac 求角am=an已知如图ab=ac ad=ae ab  dc相交于点m ac be相交于点n,角bab=角eac求角am=an
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真么写已知如图ab=ac ad=ae ab dc相交于点m ac be相交于点n,角bab=角eac 求角am=an已知如图ab=ac ad=ae ab dc相交于点m ac be相交于点n,角bab=角eac求角am=an
真么写已知如图ab=ac ad=ae ab dc相交于点m ac be相交于点n,角bab=角eac 求角am=an
已知如图ab=ac ad=ae ab  dc相交于点m ac be相交于点n,角bab=角eac
求角am=an

真么写已知如图ab=ac ad=ae ab dc相交于点m ac be相交于点n,角bab=角eac 求角am=an已知如图ab=ac ad=ae ab dc相交于点m ac be相交于点n,角bab=角eac求角am=an
证明:∵AB=AC,AD=AE,∠DAB=∠EAC,
∴∠DAC=∠AEB,
∴△ACD≌△ABE,
∴∠D=∠E,
又AD=AE,∠DAB=∠EAC,
∴△ADM≌△AEN.
am=an

30度

角bab=角eac这个是什么条件呀

◆题目表达不太清楚,在此设AB交CE于M,AC交BD于N.
证明:∵AB=AC;AD=AE;∠DAB=∠EAC.
∴⊿DAB≌⊿EAC(SAS),∠D=∠E.
∵∠DAB=∠EAC.
∴∠DAB-∠BAC=∠EAC-∠BAC,即∠DAN=∠EAM.
∵∠D=∠E(已证);AD=AE(已知);∠DAN=∠EAM(已证).
∴⊿DAN≌⊿EAM(AAS),AM=AN.

证明:∵AB=AC,AD=AE,∠DAB=∠EAC,
∴∠DAC=∠AEB,
∴△ACD≌△ABE,
∴∠D=∠E,
又AD=AE,∠DAB=∠EAC,
∴△ADM≌△AEN.
AM=AN