I think the proof is that AE = BD, AD = BE. There is a mistake in the given question.
See the diagram.
ABCD is a parallelogram. The circumcircle through A,B, & C intersects extended CD in E.
We know that in a circle, a chord inscribes the same angle at any point on the circumference on the same side.
Let angle AED = x. Let angle DAE = y and let angle ADB = z.
Compare the triangles ADE and BDE.
DE is the common side. angle BDE = angle AED. angle DAE = angle DBE.
So the triangles are similar and as one side is common, the triangles are congruent. Hence, AD = BE and AE = BD
Since, AD = BC, AD = BE = BC.