Let us take the circumcircles of ∆ DCE and ∆ EBF meet at point P.
We have to now show that the circumcircles of ∆ ADF and ∆ABC also pass through P, i.e., ADPF and ABPC are cylic quadrilaterals.
∠DCP = ∠DEP ...(i) (Angles in the same segment are equal)
Also, ∠DEP = ∠FBP ...(ii) (\(\because\) FBPE is a cyclic quadrilateral, ext ∠ = int. opp. ∠)
\(\therefore\) (i) and (ii) ⇒ ∠DCP = ∠FBP = ∠ABP. i.e.,
ext ∠ = int. opp. ∠ of quad. ABPC
⇒ ABPC is a cyclic quadrilateral.
For cyclic quadrilateral CDPE, int opp. ∠CDP = ext ∠PEB
Also, ∠PEB = ∠PFB (Angles in the same segment)
\(\therefore\) ∠CDP = ∠PFB ⇒ ∠ADP = ∠PFB
⇒ int. opp. ∠ = ext ∠ in cyclic quad. ADPF
\(\therefore\) The circumcircles of ∆s ADF, CDE, EFB and ABC intersect at point P.