把△ACF绕A点旋转90°,使AC和AB重合,设点F旋转到点G;
则有:△ABG ≌ △ACF ,
可得:AG = AF ,BG = CF ,∠GAB = ∠CAF ,∠ABG = ∠ACF = 45° ,
则有:∠EAG = ∠EAB+∠GAB = ∠EAB+∠CAF = 90°-∠EAF = 45° = ∠EAF ;
因为,在△EAG和△EAF中,AG = AF ,∠EAG = ∠EAF ,AE为公共边,
所以,△EAG ≌ △EAF ,
可得:EG = EF ;
因为,∠EBG = ∠ABC+∠ABG = 90° ,
所以,△EBG是直角三角形,
可得:BE²+BG² = EG² ,
即有:BE²+CF² = EF² .