Prove that $∠APC = ∠BPD.$

Question

Distinct points $C$ and $D$ are chosen on a circle with centre $O$ and diameter $AB$ such that they are on the same side of the diameter $AB$. The point $P$, which is different from $O$, is chosen on the diameter $AB$ such that the points $P, O, D, C$ are on one circle. Prove that $∠APC = ∠BPD.$

Taken from the 2018/2019 Estonian winter open olympiad

Im not sure how to solve this

Answer 1

We have $\angle APC = \angle ODC$ and $\angle BPD = \angle OCD$. It is easy to get conclusion.

EDIT: we have one more case of that D is between A and C. It is similar as above case.