1991 USAMO 第 5 题

Let $\, D \,$ be an arbitrary point on side $\, AB \,$ of a given triangle $\, ABC, \,$ and let $\, E \,$ be the interior point where $\, CD \,$ intersects the external common tangent to the incircles of triangles $\, ACD \,$ and $\, BCD$. As $\, D \,$ assumes all positions between $\, A \,$ and $\, B \,$, prove that the point $\, E \,$ traces the arc of a circle.

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[asy] size(220); defaultpen(1); pair A=(0,0), B=(220,0), C=(18.7723,118.523); pair D=(72.6,0); pair Ia=incenter(A,D,C), Ib=incenter(B,D,C); pair Ta=(24.9758,52.5775),Tb=(86.6196,67.4129); pair E=IntersectionPoint((Ta--Tb),(C--D)); path Oa=circle(Ia,inradius(A,D,C)); path Ob=circle(Ib,inradius(B,D,C)); pair Da=IP(Oa,A--B), Db=IP(Ob,A--B); draw(D--C--A--B--C); draw(Ta--Tb); draw(Oa); draw(Ob); dot(A,linewidth(4)); dot(B,linewidth(4)); dot(C,linewidth(4)); dot(D,linewidth(4)); dot(E,linewidth(4)); dot(Ta,linewidth(4)); dot(Tb,linewidth(4)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,W); label("$D$",D,S); label("$E$",E,NNE); [/asy]

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设 $\, D \,$ 为给定三角形 $\, ABC, \,$ 边 $\, AB \,$ 上的任意点，并设 $\, E \,$ 为内部点，其中 $\, CD \,$ 与三角形 $\, ACD \,$ 和 $\, BCD$ 的内切圆的外公切线相交。由于$\,D\,$假设$\,A\,$和$\,B\,$之间的所有位置，证明点$\,E\,$追踪圆弧。

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[asy] 大小(220);默认笔(1)；对 A=(0,0)，B=(220,0)，C=(18.7723,118.523)；对 D=(72.6,0);对 Ia=incenter(A,D,C), Ib=incenter(B,D,C);对 Ta=(24.9758,52.5775),Tb=(86.6196,67.4129);对 E=IntersectionPoint((Ta--Tb),(C--D));路径 Oa=圆(Ia,半径(A,D,C));路径 Ob=圆(Ib,inradius(B,D,C));对 Da=IP(Oa,A--B), Db=IP(Ob,A--B);绘制(D--C--A--B--C)；绘制(Ta--Tb)；绘制(Oa)；绘制(Ob)；点(A,线宽(4));点(B,线宽(4));点(C,线宽(4));点(D,线宽(4));点(E,线宽(4));点(Ta,线宽(4));点(Tb,线宽(4));标签(“$A$”，A，SW)；标签(“$B$”，B，SE)；标签(“$C$”，C，W)；标签(“$D$”，D，S)；标签(“$E$”，E，NNE)； [/asy]

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