Question

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。

Answer 1

∵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

Answer 2

∵AB//CD ∴∠BAE ∠DCE=180° 而∠BAE ∠ABE ∠AEB=∠BAE 2∠AEB =180° 2∠AEB=180°-∠BAE ∠DCE ∠CDE ∠CED=∠DCE 2∠CED=1 80° 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