将BC中点记做P,连结AP,设正方形边长为a,有△ABP≌△ADE,∠BAP=∠EADAB=a BP=PC=a/2,CF=a/4在△ABP和△PCF中：∠ABP=∠PCF=90°AB：PC=BP：CF=2△ABP∽△PCF ∠APB+∠FPC=∠CFP+∠FPC=90°所以∠APF=90°AP=√AB²...