三角形之内切圆

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Gergonne line
Proof:

$A'B'$ is the polar of $C,$ $AB$ is the polar of $C',$ by La Hire's theorem, $C''$ is the pole of $CC'.$ Similarly, $A''$ is the pole of $AA',$ $B''$ is the pole of $BB'.$

Since their polars $AA', BB',$ and $CC'$ are concurrent (in the Gergonne point $\rm Ge$), the poles $A'', B'',$ and $C''$ are collinear to a line (the Gergonne line, naturally.)

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kuing.cjhb.site/forum.php?mod=viewthread&tid=10310&extra=page%3D1
把4楼链接过来

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题简洁搬过来
kuing.cjhb.site/forum.php?mod=viewthread&tid=11334&extra=page%3D1

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把这也拉过来kuing.cjhb.site/forum.php?mod=viewthread&tid=5555

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kuing.cjhb.site/forum.php?mod=viewthread&tid=5196&extra= ... y%3Dviews&page=2

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题目：
   如图，直径$ AB $与弦$ CD $交于圆外一点$ P $，$ C,C_1 $关于$ AB $对称，$ E $为$ AB $与$ C_1D $交点。求证：$ \dfrac{OE}{OA}=\dfrac{BE}{PB} $
证明：
连接$ OC $，有\[ \angle CDE=\angle COE \]即$ CDOE $四点共园，由圆的割线定理得\[ PB\cdot PA=PC\cdot PD=PE\cdot PO \]展开\[ PB\cdot (PB+BE+OE+OA)=(PB+BE)\cdot (PB+BE+OE) \]整理后有\[  \dfrac{OE}{OA}=\dfrac{BE}{PB}  \]