如图,点E在直线AC上,AB平行CD,角AEF等于角B角CED=角D,求角BED的度数

AB平行CD∴∠A+∠C=180 ∵∠CED=∠D∴∠C=180-2×∠CED 同理∠A=180-2×∠AEB 将∠A∠C代入得 180-2∠AEB+180-2×∠DEC=180 整理得 2(∠AEB+∠DEC)=180 ∴∠AEB+∠DEC=90∴∠BED=180-90=90