试题分析:(1)过点C作CD∥AM,根据平行线相关定理即可; (2)利用三角形外角进行计算即可; (3)类比(2)的做法进行计算. 试题解析:(1)过点C作CD∥AM,
∵AM∥BN, ∴CD∥AM∥BN, ∴∠ACD=∠MAC, ∠BCD=∠CBN, ∴=∠ACD+∠BCD =∠MAC +∠CBN=(∠EAC+∠FBC)=, ∴=; (2)如图所示: ∵∠EAC的平分线所在直线与∠FBC平分线所在直线交于P, ∴∠CAP+∠CBP =(∠EAC+∠FBC)= ∵∠ACD=∠CAP+∠APC,∠BCD=∠CAB+∠BPC, ∴∠ACB=∠ACD+∠BCD = (∠APC+∠BPC)+ (∠CAP+∠CAB)= ∠APB+ ∴∠APB=﹣; (3)连接P5C并延长至点D, 根据题意知:∠CAP5+∠CBP5 =(∠EAC+∠FBC)= ∵∠ACD=∠CA P5+∠A P5C,∠BCD=∠CAB+∠B P5C, ∴∠ACB=∠ACD+∠BCD = (∠A P5C+∠B P5C)+ (∠CA P5+∠CAB)= ∠A P5B+ ∴∠A P5B=﹣. |