Check the picture below.
so far as we can tell AD = DF, by the perpendicular bisector of a chord theorem, which states that "The perpendicular bisector of a chord passes through the center of the circle".
now, what about BC? well, BC doesn't seem to have any congruent segment, if E were to be all the way up to the circle, then BC = CE, but that's not the case, so BC has no congruent segment.