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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 2 of 7: The circumcenter P repeats the same two angles
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∠PAD=∠CAB=90∘−α,∠PAC=∠BAD=90∘−β\angle PAD=\angle CAB=90^\circ-\alpha,\qquad \angle PAC=\angle BAD=90^\circ-\beta
Detailed analysis

Since PP is the circumcentre of △ACD\triangle ACD, PA=PC=PDPA=PC=PD. In isosceles △PAD\triangle PAD, the central angle ∠APD=2∠ACD=2α\angle APD=2\angle ACD=2\alpha (inscribed angle on arc ADAD), so its base angles are ∠PAD=∠PDA=90∘−α\angle PAD=\angle PDA=90^\circ-\alpha. In isosceles △PAC\triangle PAC, the central angle ∠APC=2∠ADC=2β\angle APC=2\angle ADC=2\beta, so ∠PAC=∠PCA=90∘−β\angle PAC=\angle PCA=90^\circ-\beta. Comparing with the previous step, ∠PAD=∠CAB\angle PAD=\angle CAB and ∠PAC=∠BAD\angle PAC=\angle BAD.