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AB, BC, AD and DF are four straight lines and their points of intersection A, B, C, D, E and F form four ∆s ADF, CDE, EBF and ABC. Show that the circumcircles of 4 ∆s intersect at the same point.

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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. 

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