※ 引述《nokol (騷人墨客)》之銘言：
: http://i.imgur.com/Alnj0r5.jpg
: 第七題，想請教各位大師如何證明~
: 感謝各位大師指點，謝謝您，謝謝。
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: Sent from JPTT on my Samsung SM-G935F.
搶時效，所以方法爛，但是可以證
分三個部分
(1)當FD在AD上：
E對FD做垂線交FD於J
∠FEJ = ∠BAD = 30
∠AJE = ∠AEJ = 60
=> ∠AEF = 30 = ∠EAF
=> AF = FE
(2)當F和E在AD的異側：
F對BD垂線交AD於K，交ED於M
∠BAD = 30 = ∠EFM
=> ∠AEF = ∠FKA = ∠MKD
∠BDE = a => ∠ADF = a - 30
做△EFD外接圓交AB於Q，交AD於R
∠DQE = ∠DFE = 60 => ∠ADQ = ∠DQE - ∠BAD = 30
=> ∠QEF = ∠QDF = ∠ADQ + ∠ADF = 30 + (a - 30) = a
=> ∠AER = ∠QEF - ∠REF = ∠QEF - ∠ADF = a - (a - 30) = 30
=> AR = RE => R在AE的中垂線上
又∠RFE = ∠RDE = 90 - ∠BDE = 90 - a
因為∠QEF = a，所以FR延長線會垂直於AE => F亦落在AE的中垂線上
=> AF = FE
(3)當FE在AD的同側：
∠AEF = ∠BDE = a
=> ∠ADF = 30 - a
做△EFD外交圓交AB於N，交AD於P
=> ∠PEA = ∠PDF + ∠AEF = (30 - a) + a = 30
=> AP = PE
又∠FPE = ∠FDE = 60
且∠PEA = 30 = ∠EAP
=> PF平分∠APE
延長PF交AB於H，
因為∠EPH = 60, ∠PEH = 30 => ∠HPE = 90
=> PH為等腰△AFE的中垂線
=> AF = FE
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