Question

求大神告知此题答案，要有过程

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Answer 1

Answer 2

20度

设∠ABE为X度，AC,BE交点为O
则∠AOB=∠EOC=140°-X°
因为∠ABC=2X

所以∠ACB=180°-40°-2X°=140°-2X°
又因为CE平分∠ACD
所以∠ACE=（180°-140°＋2X°）/2=20°+X°
所以∠E=180°-（140°-X°）-（20°+X°）=20°

Answer 3

三角形外角等于与他不相邻的两内角之和
所以 角ACD=角A+角ABC
角ABC=2角EBC
即 角ACD=角A+2角EBC ①

同理 角ECD=角E+角EBC
角ACD=2角ECD=2角E+2角EBC ②

联立二式可得出 角A=2角E
所以角E=20度

Answer 4

将AC、BE的交点设为F
∠ACD=∠A+∠ABC
∵BE平分∠ABC，CE平分∠ACD
∴2∠ACE=∠A+2∠ABE
∵∠A=40°
∴∠ACE=20°+∠ABE
∵∠BFC=∠E+∠ACE
∴∠BFC=∠E+20°+∠ABE
又∵∠BFC=40°+∠ABE
∴∠E+20°+∠ABE=40°+∠ABE
∴∠E=40°-20°=20°

Answer 5

E = 180 - EBC - BCE

EBC = ABC/2

BCE = ACB + ACE = ACB + ACD/2 = ACB + ( 180 - ACB )/2 = 90+ ACB/2

E = 180 - ABC/2 - ( 90+ ACB/2 ) = 90 - (ABC + ACB )/2 = 90 - ( 180 - A)/2 = a/2 = 20