You should also notice that since $CP$ is the diameter of the circle, angle $CBP$ is a right angle.
says the solution to this question. But what is the reasoning or proof behind the claim that the angle opposite the diameter ($\angle CBP$) must be a right angle?
I started to draw it out to gain some intuition, but I only got about as far showing that drawing a line from the center $O$ to the vertex of $CBP$ splits the inscribed triangle into two, where $\angle BOC + \angle BOP = \pi$. I'm not sure if that will lead to a proof, or whether this is more easily shown with e.g. the Law of Sines.
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Best Answer
In the diagram above $A$ is the centre of the circle and $CB$ is thus the diameter. Point $D$ is an arbitrary point in the circumference.
In $\triangle ACD$ $\angle CDA = \alpha$ since $AC = AD$
Thus
$$2 \cdot \alpha + \theta = 180 ^\circ$$
In $\triangle ADB$ $\angle ADB = \beta$ since $AB = AD$
Thus
$$2 \cdot \beta + (180^\circ - \theta) = 180 ^\circ$$
Add both these together
$$2 \cdot \alpha + 2 \cdot \beta + 180 ^\circ = 360 ^\circ$$
Which we can easily rearrange to show:
$$ \alpha + \beta = 90 ^\circ $$
We have thus proved $\angle CDB$ is a right angle as required.