Prove the 3 lines are concurrent In the figure, $A$ is the midpoint of a side of a regular 18-gon. The black polygon is a regular Nonagon. $O$ is the centre. I found that $BC, EF,OA$ are always concurrent but I couldn’t prove it. Any hint or solution will be appreciated. I tried adding lines and construct some triangles but seem non of them work.
I am good at plane euclidean geometry, trigonometry (Up to Year 12)but I suspect there is a elegant euclidean geometrical proof existing.

 A: Instead of proving that those three are concurrent, we prove that $OG$ bisects $FS$ (at the point $A$).
Note that $O$ is the circumcenter of the $18-$gon.
Let's assume $\angle FOA =x$.
We have:
$$1=\frac {OG}{FG} \times\frac {BG}{OG} \times \frac {FG}{BG} \\ =\frac {\sin \angle OFG}{\sin x} \times \frac{\sin \angle GOB}{ \sin \angle OBC} \times \frac{\sin \angle CBF}{\sin \angle GFB} \\ = \frac{ \sin 30^{\circ}}{\sin x} \times \frac{\sin (40^{\circ} -x)}{\sin 50^{\circ}} \times \frac{\sin 20^{\circ}}{\sin 40^{\circ}} \\ \implies  \frac{\sin (40^{\circ} -x)}{\sin x}=\frac { 2\sin 50^{\circ} \sin 40^{\circ}}{\sin 20^{\circ}} \\ =\frac{\cos 10^{\circ}}{\sin 20^{\circ}} \\ \implies \frac{\sin (40^{\circ} -x)}{\sin x}= \frac{\frac{1}{2}}{\sin 10^{\circ}}=\frac{\sin (40^{\circ} -10^{\circ})}{\sin 10^{\circ}} \\ \implies x=10^{\circ}.$$
On the other hand, $\triangle FOS$ is an isosceles triangle, and $\angle FOS =20^{\circ}.$ Hence, we are done.


A: Let $D$ be the reflection of $C$ over $OH$, so that $BC$, $FD$, and $OH$ concur at some point $X$ (by symmetry over $OH$).  Let $G, H$ be the reflections of $E, F$ over $OA$.  Then $\triangle CEG$ and $\triangle XFH$ have parallel corresponding sides, so they are homothetic about some point $Y$, which lies on $BC$, $EF$, $GH$, and also $OA$ (by symmetry over $OA$). ∎

