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[几何] 这应该算是比较简单的等腰三角形中求角

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isee posted 2024-11-16 21:04 |Read mode
Last edited by hbghlyj 2025-5-21 23:33源自知乎提问


:如图所示,等腰三角形 ABC 中,AB = AC,$\angle BAC=70^\circ$,点 O 为三角形 ABC 内一点,且 $\angle OCB=5^\circ$,$\angle ABO=25^\circ$,则 $\angle OAC=$ ______.

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kuing posted 2024-11-16 21:21
与《撸题集》P.844 题目 6.4.13 类似,书上两种证法这题都适用。

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16 年过去了,小龙女要回来了~  posted 2024-11-16 21:33

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original poster isee posted 2024-11-16 21:32
kuing 发表于 2024-11-16 21:21
与《撸题集》P.844 题目 6.4.13 类似,书上两种证法这题都适用。
完全变成书中精灵了~~

简解:以 BC 为边向上作正三角形 BCD
trg-ag-al.png

可判定 \[\triangle DCA\cong\triangle BCO\,(ASA),\]
于是有 CO=CA,从而所求角为 65 度.
isee=freeMaths@知乎

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