已知,如图,点E在AC上,AB平行CD,角B等于角AEB,角D等于角CED,求证:BE垂直ED.

∵AB//CD
∴∠BAE+∠DCE=180°
而∠BAE+∠ABE+∠AEB=∠BAE+2∠AEB=180°
2∠AEB=180°-∠BAE
∠DCE+∠CDE+∠CED=∠DCE+2∠CED=180°
2∠CED=180°-∠DCE
则2∠AEB+2∠CED=180°-∠BAE+180°-∠DCE
2(∠AEB+∠CED)=360°-(∠BAE+∠DCE)
2(∠AEB+∠CED)=360°-180°
2(∠AEB+∠CED)=180
∠AEB+∠CED=90°
∴∠BED=180°-(∠AEB+∠CED)=180°-90°=90°
即BE⊥ED