反射三角形九点圆圆心

hbghlyj · 2023-3-15 04:55

Last edited by hbghlyj 2023-4-19 15:34$\triangle A'B'C'$是$\triangle ABC$关于平面上一点$P$的反射三角形.
过九点圆圆心$N$作$AA',BB',CC'$平行线,求证:这三条直线关于对应边反射后共点,这个点是$OP$中点,$O$为外心.

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posted  <em id="authorposton51705">2023-3-16 13:16</em>
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一条直线关于一个定点的反射线是咋定义的？<br />
一知直线关于另一条直线的反射线是咋定义的？</td></tr></table>

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<div class="quote"><blockquote><font size="2"><a href="forum.php?mod=redirect&goto=findpost&pid=51705&ptid=10452" target="_blank"><font color="#999">TSC999 发表于 2023-3-16 06:16</font></a></font><br />
一条直线关于一个定点的反射线是咋定义的？<br />
一知直线关于另一条直线的反射线是咋定义的？ ...</blockquote></div>就是點對稱 Geogebra有工具<br />
<a href="https://wiki.geogebra.org/zh/%E9%BB%9E%E5%B0%8D%E7%A8%B1_%E5%B7%A5%E5%85%B7" rel="external nofollow" target="_blank">wiki.geogebra.org/zh/點對稱_工具</a><br />
<br />
选择要反射的对象。 然后，单击一个点以指定反射中心。
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<a href="https://wiki.geogebra.org/en/Reflect_Command" rel="external nofollow" target="_blank">wiki.geogebra.org/en/Reflect_Command</a><br />
<br />
反射（<对象>，<点>）<br />
     关于给定点反射几何对象。<br />
 <br />
 反射（<对象>，<线>）<br />
     关于给定的线反射一个对象。<br />
<br />
 反射（<对象>，<圆>）<br />
     相对于圆反演几何对象。<br />
<br />
 反射（<对象>，<平面>）<br />
     关于平面反射对象。</td></tr></table>

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给定直线 AB 和定点 P，如何人工作出 AB 关于 P 点的反射线 $A'B'$？<br />

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posted  <em id="authorposton51770">2023-3-17 17:29</em>
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<div class="quote"><blockquote><font size="2"><a href="forum.php?mod=redirect&goto=findpost&pid=51769&ptid=10452" target="_blank"><font color="#999">TSC999 发表于 2023-3-17 10:27</font></a></font><br />
给定直线 AB 和定点 P，如何人工作出 AB 关于 P 点的反射线？</blockquote></div>图中标的$A',B'$都不是 AB 关于 P 点的反射啊<img src="static/image/smiley/default/shocked.gif" smilieid="6" border="0" alt="" /></td></tr></table>

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<h3 class="psth xs1"><span class="icon_ring vm"></span>Comment</h3>
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<div class="psti">如何人工作出来？不要用Geogebra 软件作。 
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posted  2023-3-17 17:32</span>
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posted  <em id="authorposton51772">2023-3-17 17:33</em>
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<div class="quote"><blockquote><font size="2"><a href="forum.php?mod=redirect&goto=findpost&pid=51770&ptid=10452" target="_blank"><font color="#999">hbghlyj 发表于 2023-3-17 17:29</font></a></font><br />
图中标的$A',B'$都不是 AB 关于 P 点的反射啊</blockquote></div> 如何人工作出来？不要用Geogebra 软件作。</td></tr></table>

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<div class="quote"><blockquote><font size="2"><a href="forum.php?mod=redirect&goto=findpost&pid=51769&ptid=10452" target="_blank"><font color="#999">TSC999 发表于 2023-3-17 10:27</font></a></font><br />
给定直线 AB 和定点 P，如何人工作出 AB 关于 P 点的反射线 $A'B'$？</blockquote></div>只要延长$AP$到$A'$使$AP=A'P$即可</td></tr></table>

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<div class="quote"><blockquote><font size="2"><a href="forum.php?mod=redirect&goto=findpost&pid=51769&ptid=10452" target="_blank"><font color="#999">TSC999 发表于 2023-3-17 10:27</font></a></font><br />
给定直线 AB 和定点 P，如何人工作出 AB 关于 P 点的反射线 $A'B'$？</blockquote></div>這裡的 “反射線” 是指 點對稱 吧<br />

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<div class="psti">这样作就不是那个图了。 
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posted  2023-3-17 17:44</span>
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<div class="quote"><blockquote><font size="2"><a href="forum.php?mod=redirect&goto=findpost&pid=51774&ptid=10452" target="_blank"><font color="#999">TSC999 发表于 2023-3-17 10:44</font></a></font><br />
这样作就不是那个图了。</blockquote></div>哪个图<img src="static/image/smiley/default/shocked.gif" smilieid="6" border="0" alt="" /></td></tr></table>

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<div class="psti">4# 楼的图。 
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posted  2023-3-18 08:50</span>
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<i class="pstatus">Last edited by hbghlyj 2023-3-18 09:56</i><div class="quote"><blockquote><font size="2"><a href="forum.php?mod=redirect&goto=findpost&pid=51770&ptid=10452" target="_blank"><font color="#999">hbghlyj 发表于 2023-3-17 10:29</font></a></font><br />
图中标的$A',B'$都不是 AB 关于 P 点的反射啊</blockquote></div>4#的图是怎么作的<img src="static/image/smiley/default/shocked.gif" smilieid="6" border="0" alt="" /><br />
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<i class="pstatus">Last edited by hbghlyj 2023-4-19 15:32</i><div class="quote"><blockquote><font size="2"><u><font color="#999">TSC999 发表于 2023-3-18 01:50</font></u></font>
4# 楼的图。</blockquote></div>4#的图好像与1#无关吧{:shocked:

Ly-lie · 2023-4-19 21:08

作$A$关于$BC$的对称点$A'$,有$AP'$与$A'P$关于$BC$对称，故$A'P$平行于$DQ$,延长$OD$交$A'P$于$E$,只需$OD=DE$,作$HE'$平行于$DN$交$A'F$于$E'$,$E'G\perp BC$于$G$,$OM\perp BC$于$M$,设$OE'$与$BC$交于$D'$,由$E'H$平行于$AF$得$$ \frac {E'G}{AK}=\frac {FE'}{A'F}=\frac {AH}{AA'}$$得$$ \frac {E'G}{AK}=\frac {2OM}{2AK} $$故$E'G=OM$,即$OD'=E'D'$,故$ ND' $平行于$ E'H $平行于$ AF $平行于$ ND $,得到$ D=D' $,进而$ E=E' $,于是$ OD=DE $,即得证.

kuing · 2023-4-19 22:06

Ly-lie 发表于 2023-4-19 21:08
作$A$关于$BC$的对称点$A'$,有$AP'$与$A'P$关于$BC$对称，故$A'P$平行于$DQ$,延长$OD$交$A'P$于$E$,只需$OD ...

由E'H$平行于$AF$ 这里少了一个 $

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