完整版初一数学平行线证明题
计算证明题1如图,在四边形ABCD中,\angle A=104^{\circ}- \angle 2, \angle ABC=76^{\circ}+\angle 2,BD \bot CD于D,EF \perp CD于F,能辨认\angle 1= \angle 2吗?试说明理由.CA1F2BCE2.如图,CD\ |AB, \angle DCB=70^{\circ}, \angle CBF=20^{\circ}, \angle EFB=130^{\circ},问直线EF与AB有怎样的位置关系,为什么?CE/F3已知:如图23,AD平分\angle BAC,点F在BD上,FE|AD交AB于G,交CA的延长线于E,求证:\angle AGE= \angle E。EAGBDFC4如图,AB \ |DE, \angle 1= \angle ACB, \angle CAB= \frac {1}{2} \angle BAD,试说明:AD||BC.ADF1BEC5已知:如图22CB \perp AB,CE平分\angle BCD,DE平分\angle CDA, \angle 1+\angle 2=90^{\circ},求证:DA \perp AB.DA---EBC6如图,已知:E、F分别是AB和CD上的点,DE、AF分别交BC于G、H,\angle A= \angle D, \angle 1= \angle 2,1求证:\angle B= \angle C.EAB2GH1DF7如图,已知:\angle AOE+\angle BEF=180^{\circ}, \angle AOE+\angle CDE=180^{\circ},求证:CD||BE。8已知:如图:\angle AHF+\angle FMD=180^{\circ},GH平分\angle AHM,MN平分\angle DMH。求证:GHIMN。E:AAH/BG·NEM3BC MDPNCD12FRDCBOEF9如图,直线AB、CD被EF所截,\angle 1= \angle 2, \angle CNF= \angle BME。求证:AAH/