1.如图1,△ABC中,BE平分∠ABC,CE平分△ABC外角∠ACD,则∠A与∠E间有何数量关系,请说明原
4个回答

1.如图3

∠A=180-∠ABC-∠ACB

=180-2∠EBC-2∠ECD

=180-2(∠EBC+∠ECD)

=180-2(180-∠E)

=2∠E-180

∠E=90+1/2∠A

2.连结CO,∠EOD=∠AOB

∠OCE+∠OEC+∠COE=180

∠OCD+∠ODC+∠COD=180

∠ACB+∠OEC+∠ODC+∠EOD=∠OCE+∠OCD+∠OEC+∠ODC+∠COE+∠COD=360

∠ACB+∠EOD=360-(∠OEC+∠ODC)

∠ACB+∠EOD=360-(90+90)

∠ACB+∠EOD=180

∠ACB+∠AOB=180

3.如图1

∠ADE=∠B+∠BAD=2∠B

∠AED=∠C+∠CAE=2∠C

∠ADE+∠AED=2(∠B+∠C)

180-∠DAE=2(180-∠BAC)

180-∠DAE=360-2∠BAC

2∠BAC=180+∠DAE

∠BAC=90+1/2∠DAE

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