∫ The straight line CD is tangent to the circle with the diameter of line segment AB at point D.

∴∠ADB=90

∴ When∠∠∠∠ ∠APB is maximum, P and D coincide.

∴∠APB=90

AB = 2，AD= 1

∴sin∠DBA=AD/AB= 1/2

∴∠DBA=30

∫P and d coincide at this time.

∴∠ABP is 30 years old.

8. If the quadrilateral OABC is a parallelogram, we can get ∠B=∠AOC, ∠AOC=2∠ADC from the theorem of the circle angle, and ∠b+∠ADC = 65438+ from the properties of the inscribed quadrilateral.

Solution: ∵ quadrilateral ABCD is a quadrilateral inscribed with a circle.

∴∠B+∠D= 180

Quadrilateral OABC is a parallelogram.

∴∠AOC=∠B.

∠ AOC = 2 ∠ D。

∴∠D=60。

Connecting OD, we can get AO=OD and co = od.

∴∠OAD=∠ODA,∠OCD=∠ODC.

∴∠OAD+∠OCD=∠ODA+∠ODC=∠D=60