如图2-1-2,直线AB.CD相交于点O,OE平分∠BOD,且∠AOC=∠AOD-80°,求∠AOE的度数

.∠AOC + ∠AOD = 180°
∠AOC=∠AOD-80°
解得：∠AOC = 50°,∠AOD = 130°
∠DOE = 1/2 * ∠BOD = 1/2 * ∠AOC = 25°
∠AOE = ∠AOD + ∠DOE = 155°