Additional:

Since Yesterday Iam Strugling With That. Does The Point O Comes Out Of The Circles. How Is ACD= AQB.
Iam Find ACD+AQB=180.  Please Can  You  Give More Clarity

Yes, point O is outside the circles.

ACD is not = to AQB.

I said ACD = AQP

With AP as a chord, angle ACP = angle AQP because they are two angles on the circumference made by the same chord.

ACP = ACD because C-P-D is a straight line.

Hence ACD = AQP

Then

QPDB is cyclic so PQB + PDB = 180.

But on the straight line ODB,  angle ODC + PDB = 180

therefore, angle ODC = angle PQB.

So we have in triangle OCD,

OCD + ODC + DOC = 180     =>    ACD + PQB + DOC = 180    =>     (AQP +PQB) + DOC = 180      =>   AQB + DOA = 180

So OAQB is cyclic.

Hope that clears it up.

Bob

Hi

Have no access to my geometry software at the moment.

Maybe joining PQ and playing around with equal angles made by same chord will work.

I will come back to this when I am able. 

Bob