AC,D,E分别在AB,AC边上,BD=CE,角DCB=角EBC,CD,BE相交O,角A=50度,求角BOC的度数三角形ABC,AB>AC,D,E分别在AB,AC边上,BD=CE,角DCB=角EBC,CD,BE相交O,角A=50度,求角BOC的度数,
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∵∠DBC＝∠EBC＝30°∴∠DOB＝∠EOC＝60°∠DOE＝120°2：∵∠A＋∠DOE＝180°∴∠ADO＋∠AEO＝180°∵∠BDC＋∠ADO＝180°,∠BEC＋∠AEO＝180°∴∠BDC＝∠CED∠BDO＝∠CEO,∠DOB＝∠EOC,OB＝OC⊿BDO≌CEO∴BD＝CE