According to Greg, the answer is D, however we can also use the property of similar triangles here. In the given figure, ABC, BDC, ADB are three similar triangles. So, <BAC=<DBC=<DBA=45. This makes BDC a 45-45-90 triangles thus giving CD as 2. Shouldnt it thus be C?
Think about it, could it be a 45-45-90 triangle though? What proof do you have?
the similar angles theorem i wrote above the screenshot
\angle BAC = \angle DBC = \angle DAB = x, where 0 < x < 90. Not sure how you concluded that x must be 45. Other than that, your work is mostly correct.
Not sure where you got that from