Question

平面内,四条线段AB,BC,CD,DA首尾顺次连接,角ABC=24度,角ADC=42度

Answer 1

⑴∠BAD和∠BCD的角平分线交于点M（如图1）,求∠AMC的大小；

1/2∠BAD+∠ABC=1/2∠BCD+ ∠AMC

1/2∠BCD+∠ADC=1/2∠BAD+∠AMC

两式相加得：2∠AMC=∠ABC+∠ADC

2∠AMC=24°+ 42°

∠AMC=33°

两式相减得：∠BAD-∠BCD=∠ADC-∠ABC=42-24=18°

⑵ 点E在BA的延长线上,∠DAE的平分线和∠BCD的平分线交于点N（如图2）,则∠ANC =______.

∠ANC+1/2∠BCD=∠ABC+∠BAD+1/2∠DAE

因为∠BAD+∠DAE=180°

1/2∠DAE=90°-1/2∠BAD

所以∠ANC+1/2∠BCD=∠ABC+∠BAD+90°-1/2∠BAD

2∠ANC=2∠ABC+2×90°+∠BAD-∠BCD

2∠ANC=2×24°+2×90°+18°

∠ANC=123°