In △ABC and △DCB,
∵ AB = DC, AC = BD and BC = BC
\({AB\over DC}={AC\over BD}={BC\over BC}=1\)
∴ By SSS (side-side-side) rule of similarity
△ABC ~ △DCB
If △ABC ~ △DCB then they're also congruent.
As a result by cpct (corresponding-part-of-congruent-triangle) we can prove that
∠ABC = ∠DCB
Since AB‖CD and BC‖AD and opposite sides are also equal therefore, ABCD can be classified as a parallelogram.
As ABCD is a parallelogram the opposite angles are equal... hence ∠ABC = ∠ADC and ∠DCB = ∠DAB
I hope I made it clear enough.
You're welcome :)