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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
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人气:268 ℃ 时间:2019-08-17 20:35:31
解答
∠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+∠CD...
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