www.3112.net > 如图,AD是△ABC的高线,AD的垂直平分线分别交AB、AC于点EF,求证∠B=1/2∠AED

如图,AD是△ABC的高线,AD的垂直平分线分别交AB、AC于点EF,求证∠B=1/2∠AED

∵ ef 垂直平分ad,设ef与ad交于点h∴ah=hd 又∵ad⊥bc ad⊥ef ef=ef∴△aeh≌△deh∴∠aeh=∠deh=∠b即∠b=1/2∠aed

证明:∵EF是AD的垂直平分线,∴AF=DF,∴∠FAD=∠FDA,∵∠FAD=∠FAC+∠CAD,∠FDA=∠B+∠BAD,∵AD平分∠BAC,∴∠BAD=∠CAD,∴∠FAC=∠B.

解∵ EF 垂直平分AD,设EF与AD交于点H∴AH=HD 又∵AD⊥BC AD⊥EF EF=EF∴△AEH≌△DEH∴∠AEH=∠DEH=∠B即∠B=1/2∠AED

∵AD的垂直平分线分别交于AB,AC于点E,F∴AE=DE∴∠BAD=∠ADE∵AD⊥BC∴∠ADB=90°∴∠B+∠BAD=90°∠BDE+∠ADE=90°∵∠BAD=∠ADE∴∠B=∠BDE∵∠AED=∠B+∠BDE∴∠AED=2∠B即∠B=∠AED

(1)如图所示:(2)四边形AEDF是菱形,理由:∵AD是△ABC的角平分线,AD的垂直平分线EF,∴AE=DE,AF=DF,∠EAD=∠FAD,∠EOA=∠FOA,∵在△AEO和△AFO中,∠EAO=∠FAOAO=AO∠AOE=∠AOF,∴△AEO≌△AFO(ASA),∴AE=AF,∴AE=DE=AF=DF,∴四边形AEDF是菱形.

证明:(1)∵EF是AD的中垂线,∴DE=AE.∴∠EAD=∠EDA.(2)∵EF为中垂线,∴FD=FA.∴∠FDA=∠FAD.∵AD平分∠BAC,∴∠FAD=∠DAC,所以∠FDA=∠DAC.∴DF∥AC.(3)∵∠EAD=∠EDA,∠EAD=∠DAC+∠CAE,∠EDA=∠B+∠BAD,∴∠DAC+∠CAE=∠B+∠BAD,∵∠FAD=∠DAC,∴∠EAC=∠B.

证明:(1)∵AD为△ABC的角平分线,∴∠1=∠2,∵AD的中垂线交AB于点E、BC的延长线于点F,∴AF=DF,∴∠4=∠DAF=∠2+∠3,∵∠4=∠1+∠B,∴∠3=∠B;(2)∵EF是AD的中垂线,∴OA=OD,∴∠2=∠ODA,∵∠4=∠DAF,∴∠3=∠ODF,∵∠3=∠B,∴∠ODF=∠B,∴OD∥AB,∴∠B+∠ODB=180°.

如图,在ABC中,AB=AC,∠A=36°,AB的垂直平分线交AC于点E,垂足为点D,连接BE,则∠EBC的度数为( ).

连接DE、DF,∵EF垂直平分AD,∴AE=DE,AF=DF,设AD与EF相交于G,在RTΔAGE与RTΔAGF中,AG=AG,∠EAG=∠FAG,∠AGE=∠AGF=90°∴ΔAGE≌ΔAGF,∴AE=AF,∴AE=AF=DE=DF,∴四边形AEDF是菱形.

1.证明:∵AD的垂直平分线AB于点F,交BC的延长线于点E∴AF=DF,AE=DE∴∠FAD=∠FDA,∠DAE=∠ADE∵AD是△ABC的角平分线∴∠BAD=∠FAD=∠DAC=∠FDA∴DF∥AC2.∵∠ADE=∠B+∠BAD∴∠B+∠BAD=∠DAE=∠DAC+∠EAC∴∠B=∠EAC

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