Geometry problem

[Steven G]

Given triangle ABC
Draw median from C to D
<CAD = <DCB = x
<ADC = 45
Find x.

I know that the median bisects AB.
I can't seem to use the fact that AD=DB to find x.
What is the key thing here?
And before SK asks, yes I did draw a diagram.
I considered law of sines. I know that the two smaller triangles have the same area.

 

[Otis]

Hi Steve. I used the common side DC in Law of Sines, after setting AD=BD=1.

Spoiler

That led to:

sin(45-x)/sin(x) = sin(x)/sin(135-x)

which I solved by testing the first case that came to mind.

[imath]\;[/imath]

 

[Steven G]

OK, I had it going correctly (eventually) but messed up somehow using the law of sines. Yes, I did use AD=CD=1.
There have been quite a few of triangle problems on this forum which I could not do. I hope that using the law of sines will take care of that.
Thanks for your help.