The circle which passes through the vertices a and c of a parallelogram OABC has a centre at O if BA produced meets the circle at D then prove that ABC=2 times of angle OCD

Question is to prove that ∠APD = ∠ADP
See the attached picture to represent the problem.  
ABCD is a parallelogram.  
=>Opposite angles are same.  
=>∠ABC = ∠ADC
=>∠BAD = ∠BCD
A circle is drawn in such a way that it passes through A, B and C.  
P is a point on circle intersecting with line CD.
ABCP is cyclic quadrilateral.  
(Circle is passing through all four vertices. A, B, C and P).
From the property of Cyclic Quadrilateral, ∠ABC + ∠APC = 180°.
(Opposite angles are supplementary)
∠APC + ∠APD = 180°. (Angle on straight line)
From above 2, we can derive  ∠ABC = ∠APD.  
∠ABC = ∠ADP = ∠APD.  
∠ADP = ∠APD.