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46 lines
2.0 KiB
46 lines
2.0 KiB
2 years ago
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($11$) 四点共圆的条件是：**圆内接四边形的对角互补**
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$\because \angle ABE=\angle AHE=90^{\circ}$
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$\therefore ABEH$四点共圆
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($12$) 同理：$\because \angle AHF= \angle ADF=90^{\circ}$
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$\therefore AHFD$四点共圆
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($13$) 很明显$QD$与$BS$不在同一个三角形中，不太好直接证明两者间的关系，我们采用的办法是旋转法，将$\triangle AQD$旋转到$AQ'B$的形式上。
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因为是旋转，所以$BQ'=QD$,问题得到转化。
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下面来研究一下$Q'B^2+BS^2=QS^2$吗？
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根据以前的证明经验，我们知道$\triangle Q'SA \cong \triangle ASQ$ ($SAS$)
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$\therefore Q'S=QS$
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问题也就转化为让我们证明$\angle Q'BS=90^{\circ}$
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$\angle Q'BS=\angle ABQ'+\angle ABS=\angle ADQ+\angle ABS=45^{\circ}+45^{\circ}=90^{\circ}$
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**证毕**
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($14$)性质：共圆的四个点所连成同侧共底的两个三角形的顶角相等
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$\angle DBC=\angle EAF=45^{\circ}$
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所以$EBAQ$四点共圆,**证毕**
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($15$) 同$14$,有$\angle EAF=\angle BDC=45^{\circ}$
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所以$ADFS$四点共圆
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($16$) 利用上面$ABEQ$四点共圆的结论，所以弧$AQ$的两个对应角$\angle ABD=\angle AEQ=45^{\circ}$
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而且已知$\angle EAF=45^{\circ}$
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$\therefore \angle AQE=90^{\circ}$
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$\therefore \triangle AEQ$是等腰直角三角形
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**证毕**
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($17$) 根据上面的结论$ASFD$四点共圆,弧$AS$对的两个角$\angle ADS=\angle AFS=45^{\circ}$
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$\therefore \triangle ASF$是等腰直角三角形
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**证毕**
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($18$,$19$)因为有上面$16,17$的结论，所以$\sqrt{2}$倍的结论也是正确的
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($20$)
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**[视频证明](https://www.ixigua.com/6830219667248775687)**
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**今日头条:武老师课堂**
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