Question

如图，把三角形abc纸片沿DE折叠，点A落在四边形bcde的内部。

若a=40,1+2的度数是多少？

Answer 1

80°
因为是折皮稿的念握轿
所以∠AED=∠A'ED ∠ADE=∠A'DE
则∠A'=40°
连接AA'
∠AA'E+∠A'AE=∠1
∠AA'D+∠A'AD=∠2
∠1+∠2=∠AA'E+∠A'AE+∠AA'D+∠A'仔肆AD
=∠A+∠A'=80°

Answer 2

80°
∵是折叠的
∴∠EAD=∠DA'E
则∠A'=40°
∠BEA′＋∠A′EA=180° ∠CDA′悉腊没＋睁纳∠A′DA=180°
∴∠A′EA＋∠A′DA＋∠1＋∠2=360°
四边行内角和=360° ∴∠A′EA＋∠A′DA＋∠A′＋∠A=360°
则∠1＋∠2=∠A′＋局携∠A
∴∠1＋∠2=40°＋40°=80°

Answer 3

△行培ADE≌△A'DE
∠ADE=∠A'DE,∠AED=∠A'ED
∠1=180°-2∠AED,∠掘带誉2=180°-2∠ADE,
∠1+∠2=360°-2(∠AED+∠ADE)=360°-2(180°-∠判段A)=360°-2(180°-40°)=80°

Answer 4

g