This is another one for my "impossible geometry" file.
If X is the midpoint of both AD and BE, then ABDE is a parallelogram, and point C cannot exist.
The congruent segments are
BX ≅ EX
AX ≅ DX
Of course, vertical angles AXB and DXE are congruent.
Then, by SAS, the congruent triangles are
ΔAXB ≅ ΔDXE
or
ΔABX ≅ ΔDEX
Selection C is close, but has points D and E swapped. The figure is misleading, which may be a contributor to the problem.
