正方形ABCD中,O是对角线AC、BD的交点,过点O作OE⊥OF分别交AB、BC于E、F两点,AE=12,CF=5,则EF的值为

假设OE1和OF1为AB,BC上的垂线
OFF1与OEE1全等
AE=AE1 + E1E
CF= CF1 - FF1
AE1=CF1
E1E=FF1
AE1 = (AE+CF)/2 = 8.5
FF1 = 3.5
OF = OE = sqrt(8.5^2 + 3.5^2)
EF = sqrt(2)*OF = 13