This can be solved using the Sine Rule. Let y=AD=DB.
Note that CD is common to triangles ACD and BCD. Let z=CD
Also note that CD̂A is the exterior angle to triangle BCD, so ∠B=30°, and sin30=0.5.
sinB/CD=sinDĈB/DB, that is, 0.5/z=sin15/y, from which y=2zsin15.
CÂD=135-x.
sinCÂD/CD=sinAĈD/AD, that is, sin(135-x)/z=sin(x)/y.
ysin(135-x)=zsin(x); but y=2zsin15, so 2zsin15sin(135-x)=zsin(x), and the z's cancel.
2sin15sin(135-x)=sin(x),
2sin15(sin135cos(x)-cos135sin(x))=sin(x), sin135=sin45=√2/2, cos135=-cos45=-√2/2;
√2sin15(cos(x)+sin(x))=sin(x), now divide through by cos(x):
√2sin15(1+tan(x))=tan(x),
√2sin15+√2sin15tan(x)=tan(x),
tan(x)(1-√2sin15)=√2sin15,
tan(x)=√2sin15/(1-√2sin15)=√3/3, making x=30°. (Note that sin15=(√6-√2)/4.)
This may be one of many different ways to solve this problem, particularly because x turned out to be a "nice" angle.