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[几何] 求比值

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lrh2006 posted 2022-4-5 14:05 |Read mode
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战巡 posted 2022-4-5 22:30
回复 1# lrh2006


连$PF$,显然由于$\angle FCP=90\du$,有$PF$为直径,然后$\angle FEP=90\du$

然后$\angle EDC=\angle GFE$,再加上那些垂直,易证$\Delta GCP\sim\Delta GFE\sim \Delta CDE$,于是有
\[\begin{cases}\frac{DE}{EF}=\frac{CE}{GE}\\\frac{GP}{CD}=\frac{CG}{CE}\end{cases}\]
这样就有
\[\frac{GP}{EF}=\frac{CG\cdot CD}{GE\cdot DE}\]
这里面$\frac{CD}{DE}=\frac{1}{\sin(\angle DCE)}=\frac{1}{\sin(\angle G)}$,也就有
\[\frac{GP}{EF}=\frac{CG}{GE\sin(\angle G)}\]
按正弦定理,会有
\[\frac{GE}{CE}=\frac{\sin(\angle GCE)}{\sin(\angle G)}\]
\[GE\sin(\angle G)=CE\sin(\angle GCE)\]
其中$CG=CB=\sqrt{17^2-15^2}=8$,$CE=\frac{15\cdot 8}{17}=\frac{120}{17}$,而
\[\sin(\angle GCE)=\sin(90\du+\angle BCE)=\cos(\angle BCE)=\frac{15}{17}\]
故此
\[\frac{GP}{EF}=\frac{8}{\frac{120}{17}\cdot\frac{15}{17}}=\frac{289}{225}\]

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original poster lrh2006 posted 2022-4-6 15:42
回复 2# 战巡

懂了,谢谢老师。忘了说这是初三的题目,能不能不用正弦定理诱导公式?对不起,我的要求是不是太多了

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