Circle theorem/triange angle question I am doing practise papers and there is one question I cannot understand even with the mark scheme. I have added the pictures below:

Question (with added annotations):


Mark scheme:


The question is to find angle x (23 degrees, apparently).

At first I thought that angle OCD might be 69 degrees because of the alternate angle theorem, but I realised that it wasn't on a tangent so that wouldn't work. Other than that I have no idea - could someone please explain (in detail if possible, maths isn't my strength!) the steps to find it? Thanks in advance!
Steps hinted (justify/prove each step in the following):

The triangle $\;\Delta OCD\;$ is isosceles (why?), and thus $\;\angle OCD=\angle ODC\;$, and since $\;\angle DOC\;$ is the other angle , it must be that

$$111^\circ-x^\circ=\angle DOC=180^\circ-2\angle OCD$$

But $\;\angle OCD\;$ is an exterior one to $\;\Delta OCE\;$ and thus $\;\angle OCD=2x^\circ\;$ (why?), so we get

$$111^\circ-x^\circ=180-4x^\circ\implies3x^\circ=69^\circ\implies x=23^\circ$$
To get the $2x$:
$180-(180-2x)$ is obviously $2x$ so $OCD = 2x$ and $CDO = 2x$ because angles in the same segment are equal (the segment being the semcircle).