this means that:
AC = FD and AF = CD and CB = DE and CD = BE because opposite sides of parallelograms are congruent to each other.
since AF = CD and CD = BE, this means that AF = BE because of the transitive property that states that if a = b and b = c, then a = c.
you are given that AB congruent to BE.
AB is congruent to AC and CB by the partition postulate that states that the whole is equal to the sum of its parts.
this means that AC and CB are congruent to BE because of the transitive postulate that states if a = b and b = c, then a = c. a in this case is AC + CB which is equal to AB which is equal to BE.
AC plus CB is congruent to FD + DE because of the addition postulate that states that if a = b and c = d, then a + c = b + d.
this also means that AB is congruent to FE because of the transitive property.
this also means that AB is congruent to FE is congruent to BE because of the transitive property again.
AF is congruent to BE as stated above because they are both congruent to CD and are therefore congruent to each other based on the transitive property that states that if a = b and b = c, then a = c.
you now have all 4 sides of the combined figure congruent to each other which is the basic definition of a rhombus.
all of the other properties of a rhombus can be proven from that.
here's a compilation of basic properties and postulates and theorems that you can reference.
https://sites.google.com/site/johnnothdurft/salesianum-school/geometry-p-2/geometry-properties-theorems-postulates-etc
i don't think the partition postulate is in there, but it is in other sources that i referenced.
while not an exact word for word proof, this should be sufficient for you to get a good idea of how to prove that ABEF is a thombus.
a picture of the rhombus is shown below:
