如图,在△ABC中,AB=AC,D,E分别是BC,AC边上的点,AE=AD,且∠BAD=20°,求∠CDE的度数.
人气:127 ℃ 时间:2019-11-09 06:21:40
解答
∠CDE=10°
因为AB=AC,则∠B=∠C
因为AE=AD,则∠ADE=∠AED
∠ADE+∠CDE=∠B+∠BAD(外角定理)
等量代换∠AED+∠CDE=∠C+∠BAD
∠AED=∠C+∠CDE(外角定理)
等量代换∠C+∠CDE+∠CDE=∠C+∠BAD
2∠CDE=∠BAD=20°
则∠CDE=10°
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