(1)证明:延长AE交BC的延长线于点G. …(1分)
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/f31fbe096b63f6243793e4b28444ebf81b4ca3b1?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∵四边形ABCD是正方形,
∴AD∥CG,∠D=∠BCD=∠DCG,
∴∠DAE=∠G
∵∠FAE=∠EAD,
∴∠FAE=∠G
∴AF=FG …(3分)
∵E是DC的中点
∴DE=EC,
∵∠AED=∠GEC(对顶角相等)
∵∠D=∠ECG=90°,
∴△ADE≌△GCE (ASA)
∴AE=EG,
∴EF⊥AE. …(5分)
(2)解:若将(1)中的“正方形”改为“矩形”、“菱形”和“任意平行四边形”,其它条件不变,结论“EF⊥AE”仍然成立.
例如:“任意平行四边形”…(6分)
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/1b4c510fd9f9d72a04aacdacd72a2834359bbbb1?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
如图,延长AE交BC的延长线于G,
∵AD∥BC,E是DC的中点,
∴DE=CE,∠ADC=∠ECG,
∴∠DAE=∠G,
∴△ADE≌△GCE,
∴AE=EG,
同(1)一样可得△AFG是等腰三角形,
∴FE⊥AE.…(9分)