设正方形ABCD的边长 AB =a  ,则：
AC=a√2, CE=AC=a√2.
△ADF∽△FCE,AB/(DC-FC)=CE/FC,a/(a-FC)=a√2/FC,  FC=(a√2)/(1+√2).
∠EFC的正切值=CE/FC=(a√2)/[(a√2)/(1+√2)]=1+√2.