对李雨明的一个题目的证明
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今天李雨明老师在《我们爱几何》微信群里命制了如下题目：
问题：如图，$O$是$\triangle ABC$外心，$OD\perp BC$，$E$在$\odot(AOD)$上，$F$是$OE$与$AD$的交点，$A’$是$A$关于$\triangle FBC$的等角共轭点. 求证：$EA’$过只与$A,B,C,D$有关的定点.
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稍加观察我们可以发现定点应该就是$\odot(ABC)$与$\odot(AOD)$的第二交点，这诱使我去考察$A’$的轨迹，而后考虑射影对应.
Proof. 设$A$关于$BC$的对称点为$A_1$，$\odot(ABC)$与$\odot(AOD)$的第二交点为$X$，$AD$交$BC$于$F_1$，$A$关于$\triangle DBC$的等角共轭点为$A_2$，$AX$交$BC$于$A_3$. 则由于$\measuredangle A_1BC=\measuredangle CBA$，$\measuredangle A_2BC=\measuredangle DBA$，$\measuredangle A’BC=\measuredangle FBA$，
\[\begin{aligned}&B[A_1,A_2;C,A']\\=&B[F_1,D;A,F]\\=&[F_1,D;A,F],\end{aligned}\]同理，
\[\begin{aligned}&C[A_1,A_2;B,A']\\=&C[F_1,D;A,F]\\=&[F_1,D;A,F],\end{aligned}\]故$A_1,A_2,A’$共线.
由$OA=OX$知$DO$是$\angle ADX$的外角平分线，于是
\[\begin{aligned}&\measuredangle DAA_1\\=&\measuredangle ADO\\=&\measuredangle ODX\\=&\measuredangle OAX,\end{aligned}\]故由$AO$与$AA_1$是$\angle BAC$的等角线知$AD$与$AX$亦是$\angle BAC$的等角线，故$E$趋于$A$时，$F$沿$AD$趋于$A$，即$A’$趋于$A_3$，故$A_3$亦在$A_1A_2$上. 由$\measuredangle ODX=\measuredangle ADO=\measuredangle ODA_2$可知$D,X,A_2$共线.
考虑以$A$为中心，$AB\cdot AC$为幂，对称轴为$\angle BAC$内角平分线的轴反演反射$\varphi$，则$\varphi(BC)=\odot(ABC)$，$\varphi(O)=A_1$，$\varphi(F_1)=X$，于是$\triangle AOF_1\stackrel{+}{\sim}\triangle AXA_1$，故$\measuredangle(OF_1,XA_1)=\measuredangle OAX$，即$OF_1$与$A_1X$的交点$E_1$在$\odot(AOD)$上.
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于是
\[\begin{aligned}&X[E_1,D;A,E]\\=&O[E_1,D;A,E]\\=&[F_1,D;A,F]\\=&[A_1,A_2;A_3,A']\\=&X[A_1,A_2;A_3,A']\\=&X[E_1,D;A,A'],\end{aligned}\]这就说明$X,E,A’$共线. $\quad\Box$