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We show $ C = D $. Consider the points $ B, E $ which lie on $C$. They must also lie on $ D $, being the image of $A, B$. But (viewing from non-$O$ side) :

because of the common chord length $ |AB| = |BE| $, and hence common subtended angle size within $C$.

We show $ C = D $. Firstly note that although $C$ is a smaller radius circle than $C_{1,2}$, chords $AB$ and $BE$ can't be diameters of $C$ because that would imply $A = E$ - a contradiction - so the orientations below are well-defined. Consider the points $ B, E $ which lie on $C$. They must also lie on $ D $, being the image of $A, B$. But (viewing from non-$O$ side) :

because the non-diametrical equal length chords $AB$ and $BE$ of $C$ subtend the same angle within $C$, and these chords lie to either side of point $B$ by virtue of $A \neq E$.

Let $C_{1}$ and $C_{2}$ intersect along a diameter $ BF $, as shown in Fig 4.

Since $ B $ lies on $C_{2}$, it must have been mapped from some point $ A $ on $C_{1}$. Assume $ A \neq F $ (otherwise the proof follows from Lemma 4), and $ A \neq B $ (otherwise the proof follows from Lemma 1). $ A $ is shown on the left of $ BF $ in Fig 4 - if it was on the right, we could rotate the diagram around $ 180^\circ $ about $ BF $ so $ A $ is on the left. $E$ is shown above $C_{1}$ in Fig 4, but the same argument as below applies if it is below. The dihedral angle $ \delta \in (0, 180^\circ) $.

$ B $ also lies on $C_{1}$ so it maps to some point $ E $ on $C_{2}$. So from the rigid body motion of $C_{1}$, angle $\angle B \widehat{O} E = \Omega$, and $\mbox{chord } |AB| = \mbox{chord } |BE| $ (on $C_{1}$, $C_{2}$ respectively).

Let $C_{1}$ and $C_{2}$ intersect along a diameter $ BF $ (the 'line of nodes'), as shown in Fig 4.

Since $ B $ lies on $C_{2}$, it must have been mapped from some point $ A $ on $C_{1}$. Assume $ A \neq F $ (otherwise the proof follows from Lemma 4), and $ A \neq B $ (otherwise the proof follows from Lemma 1). $ A $ is shown on the left of $ BF $ in Fig 4 - if it was on the right, we could rotate the diagram around $ 180^\circ $ about $ BF $ so $ A $ is on the left. The dihedral angle $ \delta \in (0, 180^\circ) $.

$ B $ also lies on $C_{1}$ so it maps to some point $ E $ on $C_{2}$. So from the rigid body motion of $C_{1}$, angle $\angle B \widehat{O} E = \Omega$, and $\mbox{chord } |AB| = \mbox{chord } |BE| $ (on $C_{1}$, $C_{2}$ respectively). $E$ is shown above $C_{1}$ in Fig 4, but the same argument as below applies if it is below.

Since $ B $ lies on $C_{2}$, it must have been mapped from some point $ A $ on $C_{1}$. Assume $ A \neq F $ (otherwise the proof follows from Lemma 4), and $ A \neq B $ (otherwise the proof follows from Lemma 1). $ A $ is shown on the left of $ BF $ in Fig 4 - if it was on the right, we could rotate the diagram around $ 180^\circ $ about $ BF $ so $ A $ is on the left. The dihedral angle $ \delta \in (0, 180^\circ) $.

Since $ B $ lies on $C_{2}$, it must have been mapped from some point $ A $ on $C_{1}$. Assume $ A \neq F $ (otherwise the proof follows from Lemma 4), and $ A \neq B $ (otherwise the proof follows from Lemma 1). $ A $ is shown on the left of $ BF $ in Fig 4 - if it was on the right, we could rotate the diagram around $ 180^\circ $ about $ BF $ so $ A $ is on the left. $E$ is shown above $C_{1}$ in Fig 4, but the same argument as below applies if it is below. The dihedral angle $ \delta \in (0, 180^\circ) $.