MathLabs

Problem 4

Triangle ABCABC has circumcircle Ω\Omega and circumcenter OO. A circle Γ\Gamma with center AA intersects the segment BCBC at points DD and EE, such that BB, DD, EE, CC are all different and lie on line BCBC in this order. Let FF and GG be the points of intersection of Γ\Gamma and Ω\Omega, such that AA, FF, BB, CC, GG lie on Ω\Omega in this order. Let KK be the second intersection point of the circumcircle of triangle BDFBDF with segment ABAB, and let LL be the second intersection point of the circumcircle of triangle CGECGE with segment ACAC. Suppose that the lines FKFK and GLGL are distinct and intersect at the point XX. Prove that XX lies on the line AOAO.
Step 5 of 5: Finish the angle chase
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In plain words

The homothety turns the angle at FF into a matching angle at GG after routing through the equal base angles ∠ABF=∠GCA\angle ABF=\angle GCA of the isosceles triangle AFGAFG.

∠GFK=∠LGF\angle GFK=\angle LGF
Detailed analysis

Chasing directed angles through the homothety and the cyclic quadrilaterals: ∠GFK=∠F2BK=∠F2BF−∠ABF=∠F2DF−∠ABF=∠F2DF−∠GCA=∠GCG2−∠GCA=∠LCG2=∠LGF\angle GFK=\angle F_2BK=\angle F_2BF-\angle ABF=\angle F_2DF-\angle ABF=\angle F_2DF-\angle GCA=\angle GCG_2-\angle GCA=\angle LCG_2=\angle LGF, where ∠ABF=∠GCA\angle ABF=\angle GCA because AF=AGAF=AG makes △AFG\triangle AFG isosceles, so the two base-angle-related inscribed angles ∠ABF\angle ABF (subtending AFAF) and ∠GCA\angle GCA (subtending AGAG) are equal. This gives ∠GFK=∠LGF\angle GFK=\angle LGF, i.e. ∠KFG=∠LGF\angle KFG=\angle LGF, which by Step 1 shows XF=XGXF=XG, so XX lies on line AOAO.