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

2025-06-29 10:30:43
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回答1:

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