68 lines
2.2 KiB
Markdown
68 lines
2.2 KiB
Markdown
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将$\triangle AFD$ 旋转到$ABF'$,因为$\angle D=90^{\circ}$,所以$F',B,E,C$共线。
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因为是旋转得到$\triangle AF'B$,所以$\triangle AF'B \cong ADF$
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所以$DF=BF',\angle DAF=\angle BAF'$
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双因为$\angle BAE +\angle FAD=45^{\circ}$
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所以$\angle F'AB+\angle BAE=\angle F'AE=45^{\circ}$
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$AF'=AF,\angle F'AE=\angle EAF,AE=AE$ 根据$ASA$，所以$\triangle F'AE \cong \triangle EAF$ ①**证毕**
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$BE+DF=BE+BF'$ 旋转得到
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$BE+BF'=EF$ ②**证毕**
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$EF=EF'=BE+BF'=BE+FD$
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$\therefore C_{\triangle CEF}=CE+CF+EF=CE+CF+BE+FD=BC+CD=2AB$ ③**证毕**
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----
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因为$\triangle AF'E \cong \triangle AEF$
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$\therefore \angle 1=\angle 2$
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$\therefore \triangle BAE \cong EAH$ ⑥**证毕**
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$\therefore \angle 3=\angle 4$
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因此$AE$是$\angle BAH$平分线，④**证毕**
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**求证**：$\angle 5=\angle 7$
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$\because \angle 4+\angle 5=45^{\circ}$
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$\because \angle 3=\angle 4$
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$\therefore \angle 3+ \angle 5=45^{\circ}$
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$\because \angle 7 + \angle 3=90^{\circ}-\angle EAF=90^{\circ}-45^{\circ}=45^{\circ}$
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$\therefore \angle 5=\angle 7$
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所以,$AF$是$\angle HAD$平分线，⑤**证毕**
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$\therefore \triangle ADF \cong \triangle AHF$
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⑦ **证毕**
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---
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求证:$MA=MF$
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因为$\angle 1=\angle 2$ 已经证明的全等三角形
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$\angle 1=\angle 8$ 平行边内错角相等
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$\therefore \angle 2=\angle 8$
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$\triangle ANE$是等腰三角形，$AN=NE$ ⑨ **证毕**
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$\because \triangle AHF \cong \triangle ADF$
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$\therefore \angle HFA=\angle 9$
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$\because \angle 9=\angle MAF$ 平行线内错角相等
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$\therefore \angle HFA=\angle MAF$
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所以$\triangle MAF$是一个等腰三角形
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$\therefore MF=MA$ ⑧**证比**
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⑩ 一看就是 $12345$模型了，因为$tan \alpha=\frac{1}{2},\alpha+\beta=45^{\circ}$
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$\therefore tan \beta=\frac{1}{3}$
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**证毕**,忘记怎么证明的时候，回去复习一下 $12345$模型即可。 |