Question

如图,已知△ABC中,∠ABC=40° ,AD平分∠BAC,E为直线BC上一点,且∠EAD=∠EDA,1.若∠BAC=3

Answer 1

1、AD平分∠BAC,∠DAC=15°

设∠E=X,∠EAC=Y,则∠EAD=∠EDA = 15° + Y

则∠EAD+∠EDA+∠E = 2(15° + Y) + X =180°,即 2Y+X = 150°

∠E+ ∠ABC + ∠BAE = X+40°+30°+Y = 180°,即 X+Y =110°,解得X=70°,Y=40°

即若∠BAC=30°,则∠E=70°

2、∠CAE + ∠ABC = 180°

证明：设∠BAC= 2α,则 ∠BAD=∠DAC= α

∠ACB= 180° - ∠ABC-∠CAB = 140° - 2α,则

∠ADE=∠DAC+∠ACB = 140° - α,即∠DAE=140° - α

则∠CAE = ∠DAE+∠DAC = 140° ,又∠ABC=40°

则∠CAE+∠ABC = 180°