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
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.