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\triangle AFD 旋转到ABF',因为\angle D=90^{\circ},所以F',B,E,C共线。

因为是旋转得到\triangle AF'B,所以\triangle AF'B \cong ADF

所以DF=BF',\angle DAF=\angle BAF' 双因为\angle BAE +\angle FAD=45^{\circ} 所以\angle F'AB+\angle BAE=\angle F'AE=45^{\circ}

AF'=AF,\angle F'AE=\angle EAF,AE=AE 根据ASA,所以\triangle F'AE \cong \triangle EAF证毕

BE+DF=BE+BF' 旋转得到 BE+BF'=EF证毕

EF=EF'=BE+BF'=BE+FD \therefore C_{\triangle CEF}=CE+CF+EF=CE+CF+BE+FD=BC+CD=2AB证毕


因为\triangle AF'E \cong \triangle AEF \therefore \angle 1=\angle 2 \therefore \triangle BAE \cong EAH证毕

\therefore \angle 3=\angle 4 因此AE\angle BAH平分线,④证毕

求证\angle 5=\angle 7 \because \angle 4+\angle 5=45^{\circ} \because \angle 3=\angle 4 \therefore \angle 3+ \angle 5=45^{\circ}

\because \angle 7 + \angle 3=90^{\circ}-\angle EAF=90^{\circ}-45^{\circ}=45^{\circ}

\therefore \angle 5=\angle 7

所以,AF\angle HAD平分线,⑤证毕

\therefore \triangle ADF \cong \triangle AHF证毕


求证:MA=MF

因为\angle 1=\angle 2 已经证明的全等三角形 \angle 1=\angle 8 平行边内错角相等 \therefore \angle 2=\angle 8 \triangle ANE是等腰三角形,AN=NE证毕

\because \triangle AHF \cong \triangle ADF \therefore \angle HFA=\angle 9

\because \angle 9=\angle MAF 平行线内错角相等

\therefore \angle HFA=\angle MAF

所以\triangle MAF是一个等腰三角形 \therefore MF=MA证比

⑩ 一看就是 12345模型了,因为tan \alpha=\frac{1}{2},\alpha+\beta=45^{\circ} \therefore tan \beta=\frac{1}{3} 证毕,忘记怎么证明的时候,回去复习一下 12345模型即可。