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°