Equality of two angles in a couple of secant circumferences 
Two circumferences are secant in $A$ and $B$.
From $A$ draw a generic straight line that meets the two circumferences respectively at points $C$ and $D$. Show that the angle $\angle{CBD}$ does not change if we change the straight line through $A$ and is congruent to the angle that we get by joining $B$ and the centers of the two circles.

Now, if we draw segment $AB$ we see that however we pick $C$ and $D$ (and thus a line through $A$, since these two points identify such a line uniquely) the measure of these two angles must always be the same since they are, in their respective circumferences, angles that both subtend the same arc. That $\angle{CBD}$ is always the same then follows from the fact that in a triangle the sum of internal angles is always $180^\circ$, applied to $\Delta BCD$.
Now, it remains to show that $\angle OBO'\cong \angle CBD$ but I haven't been able to make progress on this front (using only classical euclidean geometry, so without using vectors and/or coordinates) so I would appreciate an hint about how to prove this step. Thanks. 
 A: This question should be placed in a slightly larger context.
Have a look at the following (Geogebra made) figure:

illustrating the following
Proposition :
Let two circles $\Gamma$ and $\Gamma'$ with points of intersection $P$ and $Q$. Consider the family of triangles $PMM'$ where $M$ and $M'$ are the points of intersection with $\Gamma$ and $\Gamma'$ resp. of any line passing through $Q$. All such triangles $PMM'$ are similar to each other.
Proof :
Let us establish that (with the notations of the figure), $PAA'$ and $PBB'$ are similar. It is a plain consequence, for the blue and red angles, of the inscribed angle property, establishing the similitude property (the "AA" case of similitude of triangles).
Remark :
Reamrk: One can check separately (it would be enough for the property you are looking for) that
$$\angle APA' = \angle BPB".\tag{1}$$
Indeed, the two green angles in $Q$ being opposite, have the same value ; therefore, applying once again inscribed angle property, we have, looking at the figure "brown + green = green + brown", i.e, property (1).
The case of triangle $OCC'$ (pink triangle) becomes just a particular case of the above proposition ; the fact that $CC'$ is parallel to the center's line $OO'$ being a consequence of the midsegment property.
