Question

初二几何题！急急!十万火急！

如图,在△ABC中,∠B=∠C,∠BAD=40°,且∠ADE=∠AED,求∠CDE的度数

Answer 1

∵∠ADC是△ABD一个外角,∴∠ADC=∠B+∠BAD=∠B+40°(∵ ∠BAD=40°已知条件)

又∵∠ADC由∠ADE和∠CDE构成,∴∠ADC=∠ADE+∠CDE=AED+∠CDE(∵ 已知∠ADE=∠AED)

∵∠AED是△ECD外角, ∴∠AED=∠CDE+∠C

把 ∠AED=∠CDE+∠C代入∠ADC=AED+∠CDE得:∠ADC=2∠CDE+∠C

又∵∠ADC=∠B+40°(已证)

∴2∠CDE+∠C=∠B+40°

∵∠B=∠C

∴2∠CDE=40° ∴∠CDE=40°/2=20°

Answer 2

看图~~~详解~~~~

Answer 3

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