Prove that the triangles ${\rm OAD}, {\rm OBE}, {\rm OCF}$ has another common point beside ${\rm O}$ .

Question

Given a convex hexagon ${\rm ABCDEF}$ circumscribing a circle $({\rm O})$. Assume that ${\rm O}$ is the circumcenter of the triangle ${\rm ACE}$. I see that the circumcircles of the triangles ${\rm OAD}, {\rm OBE}, {\rm OCF}$ has another common point beside ${\rm O}$. But I can't know which this point is.. I need to the help, thanks a real lot....

Answer 1

Apply the inversion wrt the incircle $\mathcal C$ of the hexagon. Let $ABCDEF$ now denote the images of the hexagon vertices. The images of the sides of the hexagon are circles tangent to $\mathcal C$ and going through $O$, which is also the center of the circumcircle of $ACE$.

Let $P_{AB}$ denote the tangency point of $\mathcal C$ and the circle going through $A$ and $B$, and similarly for other tangency points. $A$ is the midpoint of $P_{FA} P_{AB}$, $B$ is the midpoint of $P_{AB}P_{BC}$, and similarly for other segments.

We need to prove that $AD, BE, CF$ are concurrent. I have only a very unenlightened proof for this. Wlog, assume that $\mathcal C$ is centered at the origin and has radius one and $$A = (r, 0), C = (r \cos \theta_1, r \sin \theta_1), E = (r \cos \theta_2, r \sin \theta_2).$$ This completely determines the whole configuration, and $$P_{FA} = (\cos (-\phi), \sin (-\phi)), \\ P_{AB} = (\cos \phi, \sin \phi), \\ P_{BC} = (\cos (\theta_1 - \phi), \sin (\theta_1 - \phi)), \\ P_{CD} = (\cos (\theta_1 + \phi), \sin (\theta_1 + \phi)), \\ P_{DE} = (\cos (\theta_2 - \phi), \sin (\theta_2 - \phi)), \\ P_{EF} = (\cos (\theta_2 + \phi), \sin (\theta_2 + \phi)), \\ B = (P_{AB} + P_{BC})/2, \\ D = (P_{CD} + P_{DE})/2, \\ F = (P_{EF} + P_{FA})/2, \\ \phi = \arccos r.$$ Then it can be verified that $$\begin{vmatrix} y_A - y_D & x_D - x_A & x_D y_A - x_A y_D \\ y_B - y_E & x_E - x_B & x_E y_B - x_B y_E \\ y_C - y_F & x_F - x_C & x_F y_C - x_C y_F \end{vmatrix} = 0.$$