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平面几何 `duyan|20180123|2` - 语时lab
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<h1>平面几何 `duyan|20180123|2`</h1>
<div class="read-more clearfix">
<span class="date">2018/01/23</span>
<span>posted in </span>
<span class="posted-in"><a href='problemsolving.html'>解题</a></span>
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<p><mark>TODO: 微博图床已挂, 本篇缺图, 需要找回</mark></p>
<p>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.</p>
<p><img src="https://ws2.sinaimg.cn/large/006tKfTcly1fnqjnu4ou9j30iu0h7js7.jpg" alt=""/></p>
<span id="more"></span><!-- more -->
<blockquote>
<p>Qer: duyan 20180123</p>
</blockquote>
<h2 id="toc_0">解答</h2>
<p>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 \(GZ \cdot FX = GY \cdot FY\).</p>
<p>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 \(A,B,C\).</p>
<p><img src="https://ws1.sinaimg.cn/large/006tKfTcly1fnqjnwnkb2j30iu0h70to.jpg" alt=""/></p>
<p>在 \(\triangle GDZ\) 中, 有 <br/>
\[<br/>
\dfrac{GZ}{GD} = \dfrac{\sin \angle GDZ}{\sin \angle GZD} = \dfrac{\sin (C+\alpha)}{\sin(B-\alpha)}<br/>
\]<br/>
在 \(\triangle FEX\) 中, 有<br/>
\[<br/>
\dfrac{FX}{FE} = \dfrac{\sin \angle FEX}{\sin \angle FXE} = \dfrac{\sin (B+\beta)}{\sin(C-\beta)}<br/>
\]<br/>
故<br/>
\[<br/>
\begin{aligned}<br/>
\dfrac{GZ}{GD} \cdot \dfrac{FX}{FE} <br/>
&= \dfrac{\sin (C+\alpha)}{\sin(B-\alpha)} \cdot \dfrac{\sin (B+\beta)}{\sin(C-\beta)} = \dfrac{\cos (C-\beta)}{\sin(B-\alpha)} \cdot \dfrac{\cos (B-\alpha)}{\sin(C-\beta)} \\<br/>
&= \dfrac{\cos(B-C) + \cos(B+C-90^\circ)}{\cos(B-C) - \cos(B+C-90^\circ)} \\<br/>
&= \dfrac{\cos(B-C) + \sin A}{\cos(B-C) - \sin A}<br/>
\end{aligned}<br/>
\]<br/>
下面只需证明 \(\dfrac{GY}{GD} \cdot \dfrac{FY}{FE} = \dfrac{\cos(B-C) + \sin A}{\cos(B-C) - \sin A}\).</p>
<p>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<sup id="fnref1"><a href="#fn1" rel="footnote">1</a></sup>, 有</p>
<p>\[<br/>
\begin{aligned}<br/>
\dfrac{ED}{BD} \cdot \dfrac{BC}{EC} <br/>
&= \dfrac{\sin \angle EHD}{\sin \angle BHD} \cdot \dfrac{\sin \angle BHC}{\sin \angle EHC} \\<br/>
&= \dfrac{\sin 90^\circ}{\sin (\alpha-90^\circ+C)} \cdot \dfrac{\sin (180^\circ - A)}{\sin (\beta-90^\circ+B)} \\<br/>
&= \dfrac{\sin A}{\sin (B-\alpha) \sin (C-\beta)} \\<br/>
&= \dfrac{2\sin A}{\cos (B-C) - \sin A}<br/>
\end{aligned}<br/>
\]</p>
<p>因此<br/>
\[<br/>
\begin{aligned}<br/>
\dfrac{BE}{BD} \cdot \dfrac{CE}{CE} <br/>
&= \dfrac{(BD+ED) \cdot (DE+EC)}{BD \cdot EC} \\<br/>
&= \dfrac{ED \cdot (BD+DE+EC) + BD \cdot EC}{BD \cdot EC} \\<br/>
&= \dfrac{ED \cdot BC}{BD \cdot EC} +1 \\<br/>
&= \dfrac{2\sin A}{\cos (B-C) - \sin A} + 1 \\<br/>
&= \dfrac{\cos (B-C) + \sin A}{\cos (B-C) - \sin A}<br/>
\end{aligned}<br/>
\]<br/>
证毕.</p>
<div class="footnotes">
<hr/>
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<li id="fn1">
<p>交比的相关结论可用正弦定理证明. <a href="#fnref1" rev="footnote">↩</a></p>
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