Solve Angle of Intersection: ABCD & CP-DB in ABCD Square

[fawk3s]

Homework Statement

http://img341.imageshack.us/img341/1513/omfgc.png

ABCD is a square, CP is a radius for a circle, CP and DB intersect in Q, ABP=30 degrees, find DQC.

Homework Equations

none

The Attempt at a Solution

DBP is 15 degrees, that's what I've found out. Nothing else.

Its not a homework question, I randomly found it on the internet and now I can't get peace until I get how its supposed to be solved.
Please give me a hint or something.

Thanks in advance,
fawk3s

 

[fawk3s]

Never mind guys. I figured it out. This thread can be closed/deleted.
It was actually so easy that I am imbarrased I even posted this.
But I can be retarded sometimes.

Anyway, whoever is interested in the solution:
CP=BC, because both are the radius of the circle. But because DB is the diagonal of the square, DBC=45 degrees and so PBD=90-45-30=15 degrees.
Now we get PBC=45+15=60 degrees. But because PC=CB, we get that CPB=60 degrees aswell. And now we get that PCB=180-60-60=60 degrees also.
And now we can calculate CQB=180-60-45=75 degrees.
Then we can easily get DQC=180-75=105 degrees.