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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 689.
Let A₃A∩BC=X, B₃B∩AC=Y, C₃C∩AB=ZY,A₁,C₂, lie respectively on the sides CA, AB, BC of ∆ABC such that YB, CA₁,AC₂ concur at B₃.∴By Ceva’s Theorem,(AY/YC)(CC₂/C₂B)(BA₁/A₁A) = 1∴ AY/YC = (C₂B/CC₂).(A₁A/BA₁)=[(s-a)/s].[s/(s-c)]= (s-a)/(s-c)Similarly,BZ/ZA = (s-b)/(s-a) and CX/XB = (s-c)/(s-b)∴ (AY/YC).(BZ/ZA).(CX/XB)=1Hence by Converse of Ceva’s Theorem,AA₃(X), BB₃(Y), CC₃(Z) are concurrent