When does inversion respect vector addition? A recent question inquired as to how one could characterize the solutions of the equation $\frac{1}{z_1}+\frac{1}{z_2}=\frac{1}{z_1+z_2}$ for complex $z_1,z_2$. This is trivially valid whenever exactly one is zero; otherwise, the algebraic answer given by Eric Wofsey in the linked question establishes that one needs $z_1 = z_2 e^{2\pi i/3}$ i.e. the second point is obtained from the first by rotating the complex plane by $120^\circ$.
Now, the accepted answer to the linked question is entirely algebraic. But as with most complex algebra, the question can be translated into the realm of inversive geometry:

Given points $A,B$ in the plane, let $C$ be the point such that $\overrightarrow{OA}+\overrightarrow{OB}=\overrightarrow{OC}$ where $O$ is the origin. Let $A',B',C'$ are the corresponding inverse points of $A,B,C$ with respect to some circle through $O$. Show that $\overrightarrow{OA'}+\overrightarrow{OB'}=\overrightarrow{OC'}$ only if $|OA|=|OB|$ and $\angle AOB=120^\circ$.

Can someone provide a purely geometric proof of this result?
 A: We are given circle $k$ centered at $O$ and of radius $r$. Take the four points $A, C, A', C'$. By inversion, $O \in AA'$ and $|OA| \cdot |OA'| = r^2$. Analogously, $O \in CC'$ and $|OC| \cdot |OC'| = r^2$. Since $|OA| \cdot |OA'| = r^2 = |OC| \cdot |OC'|$ the four points $A, C, A', C'$ lie on a common circle $k_A$ and this circle is orthogonal to $k$. The same is true for $B, C, B', C$ - all four of them lie on a common circle $k_B$ and $k_B$ is orthogonal to $k$. Recall that as before, $O \in BB'$.
Assume $\overrightarrow{OA} + \overrightarrow{OB} = \overrightarrow{OC}$ and $\overrightarrow{OA'} + \overrightarrow{OB'} = \overrightarrow{OC'}$. That means the quadrilaterals $OACB$ and $OA'CB'$ are parallelograms such that $AC$ is parallel to $OB$ and $|AC| = |OB|$, as well as $A'C'$ is parallel to $OB'$ and $|A'C'| = |OB'|$. 
Since $O, B$ and $B'$ are collinear, $AC$ and $A'C'$ are parallel to each other (as lines parallel to the common line $OB$). But since the quadrilateral $ACA'C'$ is inscribed in the circle $k_A$ and $AC$ is parallel to $A'C'$ this quadrilateral is an isosceles trapezoid with $|AA'| = |CC'|$ which implies that $|OA| = |OC|$. 
Analogously, since $O, A$ and $A'$ are collinear, $BC$ and $B'C'$ are parallel to each other (as lines parallel to the common line $OA$). But since the quadrilateral $BCB'C'$ is inscribed in the circle $k_B$ and $BC$ is parallel to $B'C'$ this quadrilateral is an isosceles trapezoid with $|BB'| = |CC'|$ which implies that $|OB| = |OC|$. 
From the two analogous chains of arguments above we conclude that $|OA| = |OC| = |OB|$ and since $OACB$ is a parallelogram $|OA| = |BC|$ and $|OB| = |AC|$. Putting all these equalities together leads to $|AC| = |OA| = |OC| = |OB| = | BC|$ which means that $OACB$ is in fact a rhombus made of two equilateral triangles $OAC$ and $CBO$ so $\angle \, AOB = 120^{\circ}.$ 
Conversely, if we take $OACB$ to be a rhombus with angle $\angle \, AOB = 120^{\circ}$, which by the way implies $\overrightarrow{OA} + \overrightarrow{OB} = \overrightarrow{OC}$, then $|OA| = |OC| = |OB|$ so $|OA'| = |OC'| = |OB'|$ by the inversion equations $r^2 = |OA| \cdot |OA'| = |OC| \cdot |OC'| = |OB| \cdot |OB'|$. Therefore $|AA'| = |CC'| = |BB'|$. Thus the inscribed quadrilaterals $ACA'C'$ and $BCB'C'$ are isosceles trapezoids, which means that $AC$ is parallel to $A'C'$ and $BC$ is parallel to $B'C'$. But $AC$ is parallel to $OB$ and $BC$ is parallel to $OA$, and since $A' \in OA$ and $B' \in OB'$ it follows that $A'C'$ is parallel to $OB'$ and $B'C'$ is parallel to $OA'$. The latter fact implies that $OA'C'B'$ is a parallelogram (in fact a rhombus with angles $60^{\circ}$ and $120^{\circ}$) which is equivalent to the fact that $\overrightarrow{OA'} + \overrightarrow{OB'} = \overrightarrow{OC'}$.  
A: You will quickly see that when point A lies in the circle and point B outside, then the inversion will never lead to a quadrilateral which is an x-axis mirror image of the one spanned by A and B (but with smaller or larger sides, due to the inversion, of course).
So both A and B can be assumed to be outside the circle, and A' and B' inside.
Now both quadrilaterals formed by A,B and by A',B' have to be x-axis mirror images of each other again, the latter smaller (all inside the circle), meaning they are congruent, which implies that OA and OB must have the same length.
Now for the angle between OA and OB we do need some algebra again: suppose this angle is $\phi$, and if we set OA to have length a, then the diagonal OC of the larger quadrilateral can be calculated to be $2acos(\frac{\phi}{2})$, and that of the smaller one (with side length $\frac{1}{a}$) is $\frac{2cos(\frac{\phi}{2})}{a}$, and this must be the inverse of the first one.
From this we deduce that $cos(\frac{\phi}{2}) = \frac{1}{2}$, and thus $\phi = \frac{2\pi}{3}$.
QED
