第五章 相交线与平行线,5.3 平行线的性质,略,解:EFG=90,E=35,FGH=55. GE平分FGD,ABCD, FHG=HGD=FGH=55. FHG是EFH的外角, EFB=55-35=20.,解:( 1 )过点E作EGAB,其中点G在BC的左侧. ABCD,GECD, ABE+BEG=180,DCE+CEG=180, ABE+BEG+DCE+CEG=360, ABE+E+DCE=360, 即ABE+DCE=360-E. ( 2 )F=ABF+DCF. ( 3 )ABE和DCE的角平分线BF,CF相交于点F, ABE=2ABF,DCE=2DCF, ABE+DCE=2( ABF+DCF ), 由( 1 )( 2 )的结论,可得360-E=2F, 2F+E=360, 当F=125时,250+E=360, E=110.,
