denote the circumcircles of ABD and ACF intersect at A and G then angleBGC=360-240=120 So GBEC are concyclic

Let $ \{ T \} = AE \cap BF$. A $ 60^\circ$ rotation around C will map A to F and E to B, hence AE = BF and $ \angle ATF = 60^\circ ( 1 )$. From (1) we get that the quadrilaterals ADBT, BECT and CFAT are cyclic, q.e.d. Best regards, sunken rock

we know that in a quadrilateral ABCD if ∠A+∠C=180° then we can draw a circle around ABCD. let ω_1 be the circumcircle of ABD and ω_2 be the circumcircle of ACF . ω_1 intersects ω_2 at P ∠APB=180°-60°=120° and ∠APC=180°-60°=120° so ∠BPC=120° →∠BCP+∠BEC=180° so the point P would be on the circumcircle of of BEC

Try the generalization Given triangle $ ABC$ points $ D,E,F$ outside triangle $ ABC$ are chosen such that $ <BDC+<CEA+<AFB=180^0$. Prove that cicumcircles of the triangles $ BDC,CEA,AFB$ are concurrent

Call P the second common point of the circles (BCD) and (ABF), see that $ \angle AEC +\angle CPA = 180^\circ$. Best regards, sunken rock

Let $P = (ABD) \cap (BCE)$. From angle chase we may prove $P \in (CAF)$, done.