三角形之内切圆

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可以建立一个淘专辑

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AOPS
Geometry Unbound Problems for Section 11.5

Let $I$ be the incenter of triangle $ABC$. Let $K,L$ and $M$ be the points of tangency of the incircle of $ABC$ with $AB,BC$ and $CA$, respectively. The line $t$ passes through $B$ and is parallel to $KL$. The lines $MK$ and $ML$ intersect $t$ at the points $R$ and $S$. Prove that $\angle RIS$ is acute.
Evan Chen, §2.2 IMO 1998/5

First simple proof (grobber)
The problem is equivalent to showing $BI^2 > BR \cdot BS$.
But from
\[ \triangle BRK \sim \triangle MKL \sim \triangle BLS \]
we conclude
\[ BR = t \cdot \frac{MK}{ML},
  \qquad BS = t \cdot \frac{ML}{MK} \]
where $t = BK  = BL$ is the length of the tangent from $B$.
Hence $BR \cdot BS = t^2$.
Since $BI > t$ is clear, we are done.

Second projective proof
Let $N$ be the midpoint of $\overline{KL}$, and let ray $MN$ meet the incircle again at $P$.

Note that line $\overline{RBS}$ is the polar of $N$.
By Brokard's theorem, lines $MK$ and $PL$ should thus meet the polar of $N$, so we conclude $R = \overline{MK} \cap \overline{PL}$.
Analogously, $S = \overline{ML} \cap \overline{PK}$.

Again by Brokard's theorem, $\triangle NRS$ is self-polar, so $N$ is the orthocenter of $\triangle RIS$.
Since $N$ lies between $I$ and $B$ we are done.

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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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forum.php?mod=viewthread&tid=10310&extra=page=1
把4楼链接过来

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题简洁搬过来
forum.php?mod=viewthread&tid=11334&extra=page=1

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把这也拉过来forum.php?mod=viewthread&tid=5555

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forum.php?mod=viewthread&tid=5196&ext … rby=views&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 $四点共圆，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\[  \dfrac{OE}{OA}=\dfrac{BE}{PB}  \]