Solution
For this case we can see that :
AB = DE
We also know that :
m < ABC = m < EDF
For this case we have:
a) SSS
We need to have the hypothenuse congruent or the other side congruent in both triangles
b) SAS
We just need to have BC= DF
c) AAS
We need to satisfy:
m< BCA = m< DFE
d) ASA
We need to satisfy:
me) HL
We need to satisfy:
AC = EF