如图,在△ABE中,AB=AE,AD=AC,∠BAD=∠EAC,BC、DE交于点O.
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如图,在△ABE中,AB=AE,AD=AC,∠BAD=∠EAC,BC、DE交于点O.
求证:(1)△ABC≌△AED;
(2)OB=OE.
求证:(1)△ABC≌△AED;
(2)OB=OE.
证明:(1)∵∠BAD=∠EAC,
∴∠BAD+∠DAC=∠EAC+∠DAC,
即∠BAC=∠EAD.
在△ABC和△AED中
AB=AE
∠BAC=∠EAD
AC=AD,
∴△ABC≌△AED(SAS).
(2)∵由(1)知△ABC≌△AED
∴∠ABC=∠AED,
∵AB=AE,
∴∠ABE=∠AEB,
∴∠ABE-∠ABC=∠AEB-∠AED,
∴∠OBE=∠OEB.
∴OB=OE.
∴∠BAD+∠DAC=∠EAC+∠DAC,
即∠BAC=∠EAD.
在△ABC和△AED中
AB=AE
∠BAC=∠EAD
AC=AD,
∴△ABC≌△AED(SAS).
(2)∵由(1)知△ABC≌△AED
∴∠ABC=∠AED,
∵AB=AE,
∴∠ABE=∠AEB,
∴∠ABE-∠ABC=∠AEB-∠AED,
∴∠OBE=∠OEB.
∴OB=OE.
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