题目
如图,AB∥CD,直线EF分别交AB、CD于点E、F,EG平分∠BEF交CD于点G,∠1=50°,求∠2的度数. 答案:解:∵AB∥CD(已知), ∴∠1+∠BEF=180°.(两直线平行,同旁内角互补) 又∵∠1=50°(已知), ∴∠EFB=130°. ∵EG平分∠BEF ∴∠BEG=∠BEF=65°.(角平分线定义) ∵AB∥CD(已知), ∴∠2 =∠BEG=65°.(两直线平行,内错角相等)14.--Jack looks unhappy,do you know why?--Yes.He ________ his new iPhone4 yesterday( )A.has lostB.will loseC.is losingD.lost