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例1将△ABC的顶点A沿DE折叠,如图1,当点A落在原△ABC内部时,探求∠1,∠2与∠A的数量关系.解由折叠,有∠DEF=∠AED,∠EDF=∠ADE,因为∠DEF+∠AED+∠1=180°,所以∠1=180°-(∠DEF+∠AED)=180°-2∠AED,于是∠AED=180°-∠1/2.同理∠ADE=180°-∠2/2.因为∠A+∠ADE+∠AED=180°,
Example 1 will be △ ABC vertex A along DE folded, as shown in Figure 1, when the point A falls within the original △ ABC, to explore the number of ∠1, ∠2 and 数A the number of solutions by folding, there ∠DEF = ∠AED, ∠EDF = ∠ADE, since ∠DEF + ∠AED + ∠1 = 180 °, ∠1 = 180 ° - (∠DEF + ∠AED) = 180 ° -2∠AED, so ∠AED = 180 ° -∠1 / 2. Similarly, ∠ADE = 180 ° -∠2 / 2. Because ∠A + ∠ADE + ∠AED = 180 °,