Sorry for so many edits. We also know that angle BAD is x (this is not in the picture but I remember this problem).
So from the far right triangle, you know that angle CBD=180-2x-2x.
We also know that Angle BDA=180-2x.
So angle ABD=180-(180-2x)-x= 2x-x=x.
Sine ABD=BAD=x, we have an isoceles triangle. So side AD=BD=6 (from statement 1).
Triangle BCD is also isoceles so since BD=6, BC=6. So 1 is sufficient.
2 is not sufficient since we dont have any info about the lengths of the sides.
I'm looking at the problem as I type and I don't see x (or anything) for angle BAD or (DAB).
This is problem number 117 from quantitative review (OG supplement).