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

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