Solution:

∫△ADC is a right triangle

∴CD2=AC2-AD2

AD = 3，AC=5，CD>0

∴CD=4

The first case (as shown in figure 1):

CE = t = CF = CD

∴∠FDC=∠FCE

∠∠FDC+∠ADE = 90

∴∠FCE+∠ADF=90

∠∠FCE+∠DAC = 180-∠ADC = 90

∴∠DAC=∠ADF

∴DF=AF=FC=t (equilateral)

∴AC=AF=FC=2t=5

∴t=2.5

The second situation (as shown in Figure 2):

CE = CF, and point e coincides with point d.

∴CD=CF=4

∴t=4/ 1=4

To sum up, when t=2.5 or 4, △DCF is an isosceles triangle.