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[几何] 证明两个三角形的内切圆是相切的

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hbghlyj Posted 2020-7-29 14:18 |Read mode
Last edited by 2020-7-29 17:01 Screenshot_2020_0729_130405.png
X为内切圆切点,证明:△ABX,△ACX的内切圆相切
对旁心证明类似的结论

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isee Posted 2020-7-29 14:47
回复 1# ellipse

内切时

设$AP$分别切两圆分别于$P$,$P'$,则
\begin{align*}
XP&=XP'\\
\iff AX+BX-AB&=AX+CX-AC\\
\iff 2BX-2AB&=2CX-2AC\\
\iff AB+BC-AC-2AB&=AC+BC-AB-2AC\\
\iff BC-AC-AB&=BC-AB-AC
\end{align*}

显然上式恒成立,即两圆切于点$P$.

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kuing Posted 2020-7-29 15:20
旁切圆的不成立吧……
QQ截图20200729152733.png

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isee Posted 2020-7-29 15:58
回复 3# kuing


都应该画到BC直线下,切吧。

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kuing Posted 2020-7-29 16:04
回复 4# isee

原来如此……

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kuing Posted 2020-7-29 16:16
那就是一样的了:
QQ截图20200729161522.png QQ截图20200729161543.png

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 Author| hbghlyj Posted 2020-7-29 17:01
回复 4# isee
确实。我赶紧把第二张图删了。/擦汗

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