![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/3812b31bb051f819ab95edfbd9b44aed2f73e7cb?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
AF⊥BE.
证明:∵四边形ABCD是正方形,E是AD边上的中点,
∴AE=DE,AB=CD,∠BAD=∠CDA=90°,
在△BAE和△CDE中
∵
,
∴△BAE≌△CDE(SAS),
∴∠ABE=∠DCE,
∵四边形ABCD是正方形,
∴AD=DC,∠ADB=∠CDB=45°,
∵在△ADF和△CDF中,
,
∴△ADF≌△CDF(SAS),
∴∠FAD=∠FCD,
∵∠ABE=∠DCE
∴∠ABE=∠FAD,
∵∠BAD=∠BAF+∠DAF=90°,
∴∠ABE+∠BAF=90°,
∴∠AGB=180°-90°=90°,
∴AF⊥BE.