平面几何中的圆、直线与对称变换综合构造
2019-05-12 02:12
本范例演示了平面几何中直线、圆、角平分线及对称变换的综合构造过程。通过GeoGebra工具，展示了如何利用直线、垂线、中垂线及镜像对称等几何操作，探究圆与直线相交产生的特殊点及其几何性质。该构造涵盖了基础的几何作图技巧，适合用于理解几何变换在平面图形中的应用及特殊点的生成原理。
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<expression label="text1" exp=""题：在△ABC中，作∠BAC平分线交BC于点D，作AH⊥BC于H点，AD中垂线分别交ΔABH外接圆和ΔACH外接圆于点E、F，求证：D、E、F、H四点共圆""/>
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<expression label="text2" exp=""解：设EF交AD于点G，交AH于点J，以A为反演中心，\sqrt{AD·AG}为反演半径作原图像的反演，记任一点X(X∈{A,B,C,D,E,F,G,H})经反演后至X'""/>
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<expression label="text3" exp=""则由于∠AGJ=∠AHD=90°，AG·AD=AJ·AH，从而反演后D'与G、G'与D、H'与J、J与H'重合""/>
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<expression label="text4" exp=""而且由反演性质，直线BC→⊙AD'H'，直线EF→⊙AG'J'，其中由于∠AJ'G'=90°，D'为AG'中点，则⊙AG'J'的圆心为点D'，即⊙D'过点E'、F'""/>
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<expression label="text5" exp=""且⊙ABH→直线B'H'，⊙ACH→直线C'H',由于∠AB'E'=∠AEB=180°-∠AHB=90°，则AB'⊥B'H'
同理AC'⊥C'H'""/>
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<expression label="text6" exp=""连接D'E'、D'F'、B'D'、C'D'，由于AD平分∠BAC，∠B'AD'=∠BAD=∠CAD=∠C'AD'，而A、B'、C'、D'四点共圆，则B'D'=C'D'""/>
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<expression label="text7" exp=""又D'E'=D'F'为⊙D'半径，则D、E、F、H四点共圆⇔D'、E'、F'、H'四点共圆⇔∠D'E'H'=∠D'F'H'⇔∠B'E'D'=∠C'F'D'⇔ΔB'E'D'≅ΔC'F'D'""/>
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<expression label="text8" exp=""而∠D'B'E'=∠BB'E'+∠BB'G'=90°+∠AC'D'=∠AC'D'+∠AC'F'=∠D'C'F'且由于此角大小为90°+∠AC'D'>90°为钝角""/>
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<expression label="text9" exp=""又B'D'=C'F'，D'E'=D'F'，从而满足SSA型全等判定，从而ΔB'E'D'≅ΔC'F'D'⇔D、E、F、H四点共圆，得证""/>
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