Bunuel wrote:
[b]
Note that as BE=DE the triangles BCE and CDE are congruent (all three sides are equal) --> \angle{BCE}=\angle{DCE}. So if we continue the line segment CE it will meet with the crosspoint of the diagonals, let's call it point O.
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I am not sure about this point. Is it because both BOC+BDE=ECD+DC0?