AB=AC,∠1=∠2,AD=AE,=>△ABD≌△ACE(SAS) =>∠ADB=∠AEC，BD=CE，
BD=CE,DE=BC =>平行四边形BCED ------(1)
AD=AE =>∠ADE=∠AED，又∵∠ADB=∠AEC =>∠ADB-∠ADE=∠AEC-∠AED =>∠EDB=∠DEC
平行四边形BCED =>=>∠EDB+∠DEC=180 =>∠EDB=90°---(2)
(1)(2) =>四边形BCED是矩形
BD=CE,DE=BC =>平行四边形BCED ------(1)
AD=AE =>∠ADE=∠AED，又∵∠ADB=∠AEC =>∠ADB-∠ADE=∠AEC-∠AED =>∠EDB=∠DEC
平行四边形BCED =>=>∠EDB+∠DEC=180 =>∠EDB=90°---(2)
(1)(2) =>四边形BCED是矩形
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第1个回答 2012-05-03
AB=AC,∠1=∠2,AD=AE,=>△ABD≌△ACE(SAS) =>∠ADB=∠AEC，BD=CE，
BD=CE,DE=BC =>平行四边形BCED ------(1)
AD=AE =>∠ADE=∠AED，又∵∠ADB=∠AEC =>∠ADB-∠ADE=∠AEC-∠AED =>∠EDB=∠DEC
平行四边形BCED =>=>∠EDB+∠DEC=180 =>∠EDB=90°---(2)
(1)(2) =>四边形BCED是矩形
BD=CE,DE=BC =>平行四边形BCED ------(1)
AD=AE =>∠ADE=∠AED，又∵∠ADB=∠AEC =>∠ADB-∠ADE=∠AEC-∠AED =>∠EDB=∠DEC
平行四边形BCED =>=>∠EDB+∠DEC=180 =>∠EDB=90°---(2)
(1)(2) =>四边形BCED是矩形
第2个回答 2011-05-29
做∠BAC的平分线，分别交DE,BC于F,H
则AH⊥BC,AH⊥DE且DF=FE=BH=HC
∴BD=FH=CE且垂直于BC
∴四边形BCED是矩形
则AH⊥BC,AH⊥DE且DF=FE=BH=HC
∴BD=FH=CE且垂直于BC
∴四边形BCED是矩形
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