Question

在△ABC中,∠B<∠C<90°<∠A,∠BAC和∠ABC的外角平分线AE,BD分别与BC.CA的延长线交于E.D，

若∠ABC=∠AEB，∠D=∠BAD,求∠BAC

Answer 1

132°
若∠ABC=∠AEB，∠D=∠BAD,求∠BAC
BA延长线F，DB延长线BG，AB延长线BH
AE为∠FAC角平分线
∴∠FAE=∠EAC=∠AEB +∠ABC
设∠AEB为x°
∵∠ABC=∠AEB
∴∠ABC=∠AEB=x°
∠FAE=∠EAC=2 x°
∠FAE+∠EAC=4 x°
∠FAC=∠BAD=∠D=4 x°
则∠DBA=180°-8 x°
∵BG为角平分线
∴∠CBG=∠GBH
∠ABD=∠GBH（对角相等）=180°-8 x°
∠BCA+∠D=∠CBG
3 x°+4 x°=180°-8 x°
15 x°=180°
x°=12°
∠BAC=180°-4 x°=132°

Answer 2

∠D=∠BAD=4∠ABC;
∠ABD=90-∠ABC/2;
∴在三角形ABD中:8∠ABC+90-∠ABC/2=180
∠ABC=12
∴∠BAC=48
∴∠BAC=132