Question

三角形ABC中,角BAC=90度,AD垂直BC于D ,E是AD上一点,求证角BED>角C

Answer 1

因为，∠DBE < ∠ABC ，
所以，∠BED = 90°- ∠DBE > 90°- ∠ABC = ∠C 。

Answer 2

证明：因为：根据已知条件
角BAD+角DAC=90°、角C+角DAC=90°
所以：角C=角BAD
又因为：角BED=角BAD+角ABE
所以：角BED=角C+角ABE
因为：角ABE（不等于）0
所以：角BED大于角C

P.S.：题目应该在严谨点：“E是AD上不同于A、D两点的一点”，否则结果就是角BED大于等于角C

再P.S.：这种数学证明题最好画出图来分析，标出诸如角1、角2，这样看起来简介明了
，另外，这种“非等号”的证明题也要利用相关数学原理来证明，而不是想当然耳。
再再P.S.：如果我的答案你满意，请选为最佳答案，谢谢！O(∩_∩)O

Answer 3

因为 ∠BAC+∠ABC+∠C=180° ∠BAC=90°
所以 ∠B+∠C=180°-90°=90°
因为 AD⊥BC于D，E是AD上一点
所以 ∠BDA=90°，∠BDE=90°
因为 ∠EBD+∠BDE+∠BED=180°
所以 ∠EBD+∠BED=90°
因为 ∠EBD<∠ABC
所以 ∠BED>∠C

Answer 4

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