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BALTIC-WAY 2023 Problem 11

    Let ABCA B C be a triangle and let JJ be the centre of the AA-excircle. The reflection of JJ in BCB C is KK. The points EE and FF are on BJB J and CJC J, respectively, such that ∠EAB=∠CAF=90∘\angle E A B=\angle C A F=90^{\circ}. Prove that ∠FKE+∠FJE=\angle F K E+\angle F J E= 180∘180^{\circ}.

    Remark: The AA-excircle is the circle that touches the side BCB C and the extensions of ACA C and ABA B.

    Dataset text NuminaMath olympiads. Quoted with source.

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