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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 806.
Let the intersection of the three circles be P. SinceA'B' bisects arc DP, A'C' bisects arc PF, thus ∠B'A'P = 1/2 ∠DA'P∠C'A'P = 1/2 ∠FA'PSo∠B'A'C' = 1/2 ∠DA'F = ∠BACSimilarly, ∠B'C'A' = ∠BCA∠A'B'C' = ∠ABCHence, ∆ABC ~ ∆A'B'C'
To Jacobyour statement "Let the intersection of the three circles be P" may need some explanation.
Let P be the intersection of circle A' and circle C'. A,F,P,D concyclic => ∠DPF = 180 - ∠AC,F,P,E concyclic => ∠EPF = 180 - ∠C∠DPE= 360 - (180 - ∠A) - (180 - ∠C)= ∠A + ∠C= 180 - ∠B=> B,D,P,E concyclic=> Circle C' passes through P=> Three circles concurrent at PSpecial thanks to W Fung for the proof.