![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/f703738da97739126f8299d3fb198618377ae2d8?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
解:BE
2+FC
2=EF
2.理由如下:
∵△ABC是等腰直角三角形,AB=AC,
∴∠2=∠C=45°,
把△ACF绕点A顺时针旋转90°得到△ABD,如图,则∠1=∠C=45°,BD=CF,AF=AD,∠BAD=∠CAF,∠DAF=90°,
∵∠EAF=45°,
∴∠CAF+∠BAE=45°,
∴∠BAE+∠BAD=45°,即∠EAD=45°,
∴∠EAD=∠EAF,
在△AEF和△AED中
,
∴△AEF≌△AED,
∴EF=DE,
∵∠DBE=∠1+∠2=90°,
∴BE
2+BD
2=DE
2,
∴BE
2+FC
2=EF
2.