gmatfighter12 wrote:
Bunuel wrote:
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.
.
I am not sure about this point. Is it because bothBOC+BDE=ECD+DC0?
Drawing might help:
[b] In the figure, each side of square ABCD has length 1, the length of line Segment CE is 1, and the length of
...