从$A$到直径为$BC$的圆作切线，切点连线过垂心

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Geometry Unbound Problems for Section 11.5, page 100

4. 锐角三角形$ABC$，垂心$H$。从$A$到直径为$BC$的圆作切线，切点为$M$和$N$。证明$M,H,N$共线。

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$M,N,E$在以$AO$为直径的圆上$\implies\angle AME+\angle ANE=\pi$
$AM^2=AB\cdot AF=AH\cdot AE\implies\triangle AMH\sim\triangle AEM\implies\angle AHM=\angle AME$
同理$\angle AHN=\angle ANE$.
故$\angle AHM+\angle AHN=\angle AME+\angle ANE=\pi\implies M,H,N$共线.

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zhero June 24, 2010 at 4:09 PM
Let DF and BC intersect at P. Let the intersection of the tangents at F and B be X, and let the intersection of the tangents at D and C be Y. FC and BD intersect at H, so by Pascal's theorem on FFCBBD and DDFCCB, X,Y,H, and P are collinear. The polars of X and Y are BF and CD and intersect at A. Since X, Y, and H are collinear, the polar of H must concur with the polars of X and Y, that is, the polar of H must pass through A. It follows that H lies on the polar of A, which is line MN, so we are done.