如图,直线AB,CD相交于点O,OE平分角AOC,∠AOD比∠AOE打75°,求∠BOD的度数

∠AOD比∠AOE大75°,即∠AOD=∠AOE+75°,OE为∠AOC平分线,即∠AOE=∠COE,那么∠COE+∠AOE+∠AOD=2∠AOE+∠AOE+75°=180°,得出∠AOE=35°,那么∠AOC=2∠AOE=70°,对角相等,∠BOD=∠AOC=70°.