Question

如图,把三角形ABC纸片沿DE折叠,当点A落点A'处,求证角1+角2=2角A

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Answer 1

∵把三角形ABC纸片沿DE折叠,当点A落点A'处
∴∠ADE=∠A'DE ∠AED=∠A'ED
∴∠1=180°-2∠ADE ∠2=180°-2∠AED
∴∠1+∠2=360°-2（∠ADE+∠AED）=360°-2（180°-∠A）=2∠A

Answer 2

据题意得：⊿ADE≌⊿A'DE.
∴∠ADE=∠A'DE,∠AED=A'ED.
∵∠1=180°-∠A'DE-∠ADE=180°-2∠ADE,
∠2=180°-∠A'ED-∠AED=180°-2∠AED.
∴∠1+∠2=360°-2(∠ADE+∠AED)
∵∠A=180°-(∠ADE+∠AED)
∴∠1+∠2=2∠A