三角形中∠ABC=∠ACB,D为BC上一点,E为直线AC上一点,且∠ADE=∠AED;求证:∠BAD=2∠CDE急用
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![三角形中∠ABC=∠ACB,D为BC上一点,E为直线AC上一点,且∠ADE=∠AED;求证:∠BAD=2∠CDE急用](/uploads/image/z/2505166-70-6.jpg?t=%E4%B8%89%E8%A7%92%E5%BD%A2%E4%B8%AD%E2%88%A0ABC%3D%E2%88%A0ACB%2CD%E4%B8%BABC%E4%B8%8A%E4%B8%80%E7%82%B9%2CE%E4%B8%BA%E7%9B%B4%E7%BA%BFAC%E4%B8%8A%E4%B8%80%E7%82%B9%2C%E4%B8%94%E2%88%A0ADE%3D%E2%88%A0AED%3B%E6%B1%82%E8%AF%81%3A%E2%88%A0BAD%3D2%E2%88%A0CDE%E6%80%A5%E7%94%A8)
三角形中∠ABC=∠ACB,D为BC上一点,E为直线AC上一点,且∠ADE=∠AED;求证:∠BAD=2∠CDE急用
三角形中∠ABC=∠ACB,D为BC上一点,E为直线AC上一点,且∠ADE=∠AED;求证:∠BAD=2∠CDE
急用
三角形中∠ABC=∠ACB,D为BC上一点,E为直线AC上一点,且∠ADE=∠AED;求证:∠BAD=2∠CDE急用
∠CDE+∠ACB=∠AED(因为∠ABC=∠ACB)
所以∠CDE+∠ABC=∠AED
∠ABC+∠BAD=∠ADC=∠ADE+∠EDC(因为∠ADE=∠AED)
∠ABC+∠BAD=∠AED+∠EDC=∠EDC+∠ABC+∠EDC
即∠ABC+∠BAD=∠CDE+∠ABC+∠CDE消去∠ABC
等∠BAD=∠CDE+∠CDE=2∠CDE
有图才会有真相- -。
(1)∵∠ABC=∠ACB
∴∠C=∠B
∵∠ADB+∠ADE+∠CDE=180°
∠CED+∠AED=180°
又∵∠ADE=∠AED
∴∠CED=∠ADB+∠CDE
∵∠ADB+∠B+∠BAD=180°
∠ADB+∠C+∠BAD=180°
∠CDE+∠CED+∠C=180°
∠CDE+∠ADB+∠CDE(∠CED=∠ADB...
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(1)∵∠ABC=∠ACB
∴∠C=∠B
∵∠ADB+∠ADE+∠CDE=180°
∠CED+∠AED=180°
又∵∠ADE=∠AED
∴∠CED=∠ADB+∠CDE
∵∠ADB+∠B+∠BAD=180°
∠ADB+∠C+∠BAD=180°
∠CDE+∠CED+∠C=180°
∠CDE+∠ADB+∠CDE(∠CED=∠ADB+∠CDE)+∠C=180°
∴∠BAD=2∠CDE(去掉相同的:∠ADB and ∠C)
(2)
不知道什么是反向延长线.但是第一解出来对第二也有帮助吧.....
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