证明：将三角形ABD绕点A旋转,使AB边与AC边重合,点D旋转后为点F,连接EF∵△ABD全等于△ACF∴∠CAF＝∠BAD,∠ABD＝∠ACF,CF＝BD,AF＝AD∵等腰直角三角形ABC∴∠ABC＝∠ACB＝45∴∠BCF＝∠ACB+∠ACF＝∠ACB+∠ABD＝90∴...