TODO: 微博图床已挂, 本篇缺图, 需要找回

已知 \(H\) 为 \(\triangle ABC\) 的垂心, 过 \(H\) 的直线交 \(BC, AB\) 于 \(D,Z\), 过 \(H\) 且垂直于 \(ZH\) 的另一条直线交 \(BC, AC\) 于 \(E,X\), 点 \(Y\) 使得 \(DY \parallel AC, EY \parallel AB\). 求证: \(X,Y,Z\) 三点共线.

Qer: duyan 20180123

解答

延长 \(YD\) 交 \(AB\) 于 \(G\), 延长 \(YE\) 交 \(AC\) 于 \(F\). 比较 \(\triangle GYZ\) 和 \(\triangle FXY\), 已有两边对应平行, 故只需证明它们相似即可. 下面设法证明 \(\dfrac{GZ}{GY} = \dfrac{FY}{FX}\), 亦即证明 \(GZ \cdot FX = GY \cdot FY\).

记 \(\angle HDE = \alpha\), \(\angle HED = \beta\), 有 \(\alpha + \beta = 90^\circ\). 另外, 下面直接将 \(\triangle ABC\) 的三内角简记为 \(A,B,C\).

在 \(\triangle GDZ\) 中, 有
\[
\dfrac{GZ}{GD} = \dfrac{\sin \angle GDZ}{\sin \angle GZD} = \dfrac{\sin (C+\alpha)}{\sin(B-\alpha)}
\]
在 \(\triangle FEX\) 中, 有
\[
\dfrac{FX}{FE} = \dfrac{\sin \angle FEX}{\sin \angle FXE} = \dfrac{\sin (B+\beta)}{\sin(C-\beta)}
\]
故
\[
\begin{aligned}
\dfrac{GZ}{GD} \cdot \dfrac{FX}{FE}
&= \dfrac{\sin (C+\alpha)}{\sin(B-\alpha)} \cdot \dfrac{\sin (B+\beta)}{\sin(C-\beta)} = \dfrac{\cos (C-\beta)}{\sin(B-\alpha)} \cdot \dfrac{\cos (B-\alpha)}{\sin(C-\beta)} \\
&= \dfrac{\cos(B-C) + \cos(B+C-90^\circ)}{\cos(B-C) - \cos(B+C-90^\circ)} \\
&= \dfrac{\cos(B-C) + \sin A}{\cos(B-C) - \sin A}
\end{aligned}
\]
下面只需证明 \(\dfrac{GY}{GD} \cdot \dfrac{FY}{FE} = \dfrac{\cos(B-C) + \sin A}{\cos(B-C) - \sin A}\).

显然, \(\dfrac{GY}{GD} \cdot \dfrac{FY}{FE} = \dfrac{BE}{BD} \cdot \dfrac{CE}{CE}\), 考虑 \(B,D,E,C\) 四点的交比¹, 有

\[
\begin{aligned}
\dfrac{ED}{BD} \cdot \dfrac{BC}{EC}
&= \dfrac{\sin \angle EHD}{\sin \angle BHD} \cdot \dfrac{\sin \angle BHC}{\sin \angle EHC} \\
&= \dfrac{\sin 90^\circ}{\sin (\alpha-90^\circ+C)} \cdot \dfrac{\sin (180^\circ - A)}{\sin (\beta-90^\circ+B)} \\
&= \dfrac{\sin A}{\sin (B-\alpha) \sin (C-\beta)} \\
&= \dfrac{2\sin A}{\cos (B-C) - \sin A}
\end{aligned}
\]

因此
\[
\begin{aligned}
\dfrac{BE}{BD} \cdot \dfrac{CE}{CE}
&= \dfrac{(BD+ED) \cdot (DE+EC)}{BD \cdot EC} \\
&= \dfrac{ED \cdot (BD+DE+EC) + BD \cdot EC}{BD \cdot EC} \\
&= \dfrac{ED \cdot BC}{BD \cdot EC} +1 \\
&= \dfrac{2\sin A}{\cos (B-C) - \sin A} + 1 \\
&= \dfrac{\cos (B-C) + \sin A}{\cos (B-C) - \sin A}
\end{aligned}
\]
证毕.