⑴∠BAD和∠BCD的角平分線交于點M（如圖1）,求∠AMC的大小；
1/2∠BAD+∠ABC=1/2∠BCD+ ∠AMC
1/2∠BCD+∠ADC=1/2∠BAD+∠AMC
兩式相加得：2∠AMC=∠ABC+∠ADC
2∠AMC=24°+ 42°
∠AMC=33°
兩式相減得：∠BAD-∠BCD=∠ADC-∠ABC=42-24=18°
⑵ 點E在BA的延長線上,∠DAE的平分線和∠BCD的平分線交于點N（如圖2）,則∠ANC =______.
∠ANC+1/2∠BCD=∠ABC+∠BAD+1/2∠DAE
因為∠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°