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?
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 $\overrightarrow{OA'} + \overrightarrow{OB'} = \overrightarrow{OC'}$.