MathLabs

Problem 2

Let Ω\Omega and Γ\Gamma be circles with centres MM and NN, respectively, such that the radius of Ω\Omega is less than the radius of Γ\Gamma. Suppose circles Ω\Omega and Γ\Gamma intersect at two distinct points AA and BB. Line MNMN intersects Ω\Omega at CC and Γ\Gamma at DD, such that points CC, MM, NN and DD lie on the line in that order. Let PP be the circumcentre of triangle ACDACD. Line APAP intersects Ω\Omega again at E≠AE\ne A. Line APAP intersects Γ\Gamma again at F≠AF\ne A. Let HH be the orthocentre of triangle PMNPMN. Prove that the line through HH parallel to APAP is tangent to the circumcircle of triangle BEFBEF.
Step 5 of 7: T also lies on lines ME and NF
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F,T,N collinear,E,T,M collinearF,T,N\ \text{collinear},\qquad E,T,M\ \text{collinear}
Detailed analysis

Since DF∥ACDF\parallel AC, triangles FEA′FEA' and FADFAD are related by a homothety centred at FF; this homothety sends the circumcentre TT of △FEA′\triangle FEA' to the centre of the circle through F,A,DF,A,D, which is NN (the centre of Γ\Gamma, as A,D,F∈ΓA,D,F\in\Gamma). Hence F,T,NF,T,N are collinear. By the symmetric homothety at EE (using CE∥ADCE\parallel AD), E,T,ME,T,M are collinear.