this hexagon is simply inscribed so imply Brianchon

By Ceva’s theorem, we can prove a more general statement: Let $ABCDEF$ be an inscribed hexagon. Then $AD,BC,EF$ are concurrent if and only if $AB\cdot CD\cdot EF=BC\cdot DE\cdot FA$

georgeado17 wrote: this hexagon is simply inscribed so imply Brianchon Briachon is about circumscribed

Semcio wrote: georgeado17 wrote: this hexagon is simply inscribed so imply Brianchon Briachon is about circumscribed It's both

More simply: the concurrency point is the incenter of $ACE$.

Let $O$ be the circumcircle of $ \triangle AEC $. Lemma: $F, B, D$ lie on the circumcircle of $\triangle AEC$ Proof: Draw segment $EO$ to form the triangle $EOF$. Note that the perpendicular bisector of $EF$ must necessarily pass through the circuncenter $O$ but, by properties of the circle (you can find many proofs) if a perpendicular bisector of a chord pases through the center, this bisector halves the chord. So the perpendicular bisector $EF$ cuts $EF$ by a half, but that means that $\triangle EOF$ is an isosceles triangle. $\implies$ $F$ lies on the circumcircle of $ \triangle AEC$. You can proceed similarly in the other two points. $\blacksquare$ Now by the Incenter-excenter Lemma: $AD$, $CF$ and $BE$ concur at the incenter of $\triangle AEC$ $\blacksquare$.