[几何] 一道内切圆半径的题目

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hbghlyj posted 2020-4-8 00:08 |Read mode

$\triangle ABC$边BC上一点D,$\triangle ABD$与$\triangle ACD$的内切圆半径相等,求证：\[\S{ABC}=AD^2\Leftrightarrow\angle BAC=90^\circ\]

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2025-6-8 06:54 GMT+8

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