7. 如图，直线m//n，AB⊥BC，∠1 = 35°，∠2 = 62°，则∠BCD的度数为 （ ）
A. 97° B. 117° C. 125° D. 152°
　　　　　　　　　　

B 解析：如图，过点B作$BE// m$，过点C作$CF// n$。$\because m// n$，$\therefore m// BE// CF// n$。$\therefore \angle ABE=\angle1 = 35^{\circ}$，$\angle BCF=\angle EBC$，$\angle DCF=\angle2 = 62^{\circ}$。又$\because AB\perp BC$，$\therefore \angle ABC = 90^{\circ}$。$\therefore \angle EBC=90^{\circ}-35^{\circ}=55^{\circ}$。$\therefore \angle BCF=\angle EBC = 55^{\circ}$。$\therefore \angle BCD=\angle BCF+\angle DCF=55^{\circ}+62^{\circ}=117^{\circ}$。

8. 如图，直线AE//DF，若∠ABC = 120°，∠DCB = 95°，则∠1 + ∠2 = __________.
　　　　　　

$35^{\circ}$

9. 如图，若AB//EF，∠C = 90°，则α，β与γ之间的数量关系是__________.
　　　　　　

$\alpha+\beta-\gamma = 90^{\circ}$

10. 学习了平行线的判定与性质后，某兴趣小组提出如下问题：
已知AB//CD.
（1）如图①，若∠C = 3∠B，求∠B的度数.
（2）如图②，当点E，F在两平行线之间，且位于点B，C所在直线的异侧时，求证：∠B + ∠E = ∠C + ∠F.
（3）在（2）的条件下，如图③，∠ABE = 3∠EBP，∠CFE = 3∠EFP. 若∠E = 88°，∠C = 130°，求∠BPF的度数.
　

（1）$\because AB// CD$，$\therefore \angle B+\angle C = 180^{\circ}$。$\because \angle C = 3\angle B$，$\therefore \angle B+3\angle B = 180^{\circ}$。$\therefore \angle B = 45^{\circ}$ （2）过点E向右作$EM// AB$，过点F向左作$FN// AB$。$\because AB// CD$，$\therefore AB// CD// EM// FN$。$\therefore \angle B+\angle BEF+\angle FEM = 180^{\circ}$，$\angle EFN+\angle EFC+\angle C = 180^{\circ}$，$\angle EFN=\angle FEM$。$\therefore \angle B+\angle BEF=\angle C+\angle EFC$ （3）由（2），知$\angle ABE+\angle E=\angle CFE+\angle C$，$\therefore \angle ABE-\angle CFE=\angle C-\angle E = 130^{\circ}-88^{\circ}=42^{\circ}$。$\because \angle ABE = 3\angle EBP$，$\angle CFE = 3\angle EFP$，$\therefore \angle EBP-\angle EFP = 14^{\circ}$。设BP交EF于点O。$\because \angle EBO+\angle E+\angle BOE=\angle POF+\angle EFP+\angle P = 180^{\circ}$，$\angle BOE=\angle POF$，$\angle E = 88^{\circ}$，$\therefore \angle EBO+88^{\circ}=\angle P+\angle EFP$。$\therefore \angle P=88^{\circ}+\angle EBO-\angle EFP=88^{\circ}+14^{\circ}=102^{\circ}$

11. 已知AB//CD，∠ABE与∠CDE的平分线相交于点F.
（1）如图①，若∠E = 80°，求∠BFD的度数；
（2）如图②，∠ABM = $\frac{1}{3}$∠ABF，∠CDM = $\frac{1}{3}$∠CDF，写出∠M与∠E之间的数量关系，并证明你的结论；
（3）若∠ABM = $\frac{1}{n}$∠ABF，∠CDM = $\frac{1}{n}$∠CDF，设∠E = m，直接用含m，n的式子表示∠M的度数.
　　

（1）如图，分别过点E，F作$EG// AB$，$FH// AB$。$\because AB// CD$，$\therefore EG// AB// FH// CD$。$\therefore \angle ABF=\angle BFH$，$\angle CDF=\angle DFH$，$\angle ABE+\angle BEG = 180^{\circ}$，$\angle GED+\angle CDE = 180^{\circ}$。$\therefore \angle ABE+\angle BEG+\angle GED+\angle CDE = 360^{\circ}$。$\because \angle BED=\angle BEG+\angle GED = 80^{\circ}$，$\therefore \angle ABE+\angle CDE = 280^{\circ}$。$\because \angle ABE$与$\angle CDE$的平分线相交于点F，$\therefore$易得$\angle ABF+\angle CDF = 140^{\circ}$。$\because \angle ABF=\angle BFH$，$\angle DFH=\angle CDF$，$\therefore \angle BFD=\angle BFH+\angle DFH=\angle ABF+\angle CDF = 140^{\circ}$ （2）$6\angle M+\angle E = 360^{\circ}$ $\because \angle ABM=\frac{1}{3}\angle ABF$，$\angle CDM=\frac{1}{3}\angle CDF$，$\therefore \angle ABF = 3\angle ABM$，$\angle CDF = 3\angle CDM$。$\because \angle ABE$与$\angle CDE$的平分线相交于点F，$\therefore$易得$\angle ABE = 6\angle ABM$，$\angle CDE = 6\angle CDM$。由（1），易得$\angle ABE+\angle E+\angle CDE = 360^{\circ}$，$\therefore 6\angle ABM+6\angle CDM+\angle E = 360^{\circ}$。由题意，易得$\angle M=\angle ABM+\angle CDM$，$\therefore 6\angle M+\angle E = 360^{\circ}$ （3）由（2），易得$2n\angle ABM+2n\angle CDM+\angle E = 360^{\circ}$。$\because \angle E = m$，$\angle M=\angle ABM+\angle CDM$，$\therefore \angle M=\frac{360^{\circ}-m}{2n}$