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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 3 of 7: The second intersections E and F give two parallel lines
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∠AEC=∠ABC=∠CAB=90∘−α=∠EAD ⟹ CE∥AD,DF∥AC\angle AEC=\angle ABC=\angle CAB=90^\circ-\alpha=\angle EAD\ \Longrightarrow\ CE\parallel AD,\qquad DF\parallel AC
Detailed analysis

In circle Ω\Omega, the inscribed angle ∠AEC\angle AEC subtends the same arc ACAC as ∠ABC\angle ABC, so ∠AEC=∠ABC\angle AEC=\angle ABC; since △CAB\triangle CAB is isosceles this equals ∠CAB=90∘−α\angle CAB=90^\circ-\alpha, which by the previous step equals ∠EAD\angle EAD (as EE lies on ray APAP, so ∠EAD=∠PAD\angle EAD=\angle PAD). Thus ∠AEC=∠EAD\angle AEC=\angle EAD are equal alternate angles for transversal AEAE cutting lines CECE and ADAD, forcing CE∥ADCE\parallel AD. The symmetric computation in circle Γ\Gamma (using ∠AFD=∠ABD=∠BAD=90∘−β=∠FAC\angle AFD=\angle ABD=\angle BAD=90^\circ-\beta=\angle FAC) gives DF∥ACDF\parallel AC.