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See diagram. You can see a general quadrilateral with AB || DC and AD = BC.

Extend the sides AD and BC till E and F as shown.  

As AB and CD are two parallel lines and AD intersects them both, the angles D and EAB are same.  But angle EAB = 180 - A.  so,
             angle D = 180 - A

Similarly, the line BC intersects parallel lines AB and DC, so
           angle C = angle FBA = 180 - B 

Now, draw perpendiculars from D and C onto AB meeting AB at G and H respectively.  
     Since AB || DC, the sides DG || CH.  
               Also,  DG = CH = distance between the parallel lines.

Looking at the triangles DGA and CHB, we find that
            DG = CH,    AD = BC (given),  angle G = angle H  = 90
Δ DGA and Δ CHB are congruent.  Hence angle A = angle B.
     So, angle C = 180 - angle A = 180 - angle B = angle D

2 5 2
I hope that is good enough reasoning and proof.