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 1 of 7: Line MN is the perpendicular bisector of AB
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In plain words

Reflecting the whole picture across the line of centers swaps the two circles' intersection points, so C and D see A and B symmetrically.

α:=∠ACD=∠BCD,β:=∠ADC=∠BDC\alpha:=\angle ACD=\angle BCD,\qquad \beta:=\angle ADC=\angle BDC
Detailed analysis

The common chord ABAB of two intersecting circles is always perpendicular to the line joining their centres, and that line bisects ABAB; since C,DC,D lie on line MNMN, both are equidistant from AA and BB, so CA=CBCA=CB and DA=DBDA=DB. Reflection across MNMN therefore fixes C,D,M,NC,D,M,N and swaps A↔BA\leftrightarrow B, so it carries ∠ACD\angle ACD to ∠BCD\angle BCD and ∠ADC\angle ADC to ∠BDC\angle BDC, giving the two equal pairs above; consequently ∠ACB=2α\angle ACB=2\alpha and ∠ADB=2β\angle ADB=2\beta. Since △CAB\triangle CAB and △DAB\triangle DAB are isosceles with apex angles 2α,2β2\alpha,2\beta, their base angles are ∠CAB=∠CBA=90∘−α\angle CAB=\angle CBA=90^\circ-\alpha and ∠DAB=∠DBA=90∘−β\angle DAB=\angle DBA=90^\circ-\beta.