Question

如图,在三角形ABC中,角A=60度,角ABC,角ACB的角平分线相交于点O,过点O分别作OE平行于AB,OF平行于AC

Answer 1

∵∠A等于60° ,

∴∠ABC＋∠ACB=120°,

∵BO,CO分别平分∠ABC和∠ACB,

∴∠ABO＋∠ACO=½∠ABC+∠ACB=½X120°=60°,

∵OE、OF分别平行于AB＼AC,

∴∠BOE+∠COF=∠ABO+∠AOC=60°