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2002 AMC 10A Problem 20

Problem 20 of 25HarderGeometry

Points A,A, B,B, C,C, D,D, E,E, and FF lie, in that order, on AF‾,\overline{AF}, dividing it into five segments, each of length 1.1. Point GG is not on line AF.AF. Point HH lies on GD‾,\overline{GD}, and point JJ lies on GF‾.\overline{GF}. The line segments HC‾,\overline{HC}, JE‾,\overline{JE}, and AG‾\overline{AG} are parallel. Find HCJE.\frac{HC}{JE}.

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Solution

Since HC∥AG,HC\parallel AG, △DHC∼△DGA,\triangle DHC\sim\triangle DGA, so HCAG=DCDA=13,\dfrac{HC}{AG}=\dfrac{DC}{DA}=\dfrac{1}{3}, giving HC=AG3.HC=\dfrac{AG}{3}. Since JE∥AG,JE\parallel AG, △FJE∼△FGA,\triangle FJE\sim\triangle FGA, so JEAG=FEFA=15,\dfrac{JE}{AG}=\dfrac{FE}{FA}=\dfrac{1}{5}, giving JE=AG5.JE=\dfrac{AG}{5}. Therefore HCJE=AG3AG5=53.\dfrac{HC}{JE}=\dfrac{\frac{AG}{3}}{\frac{AG}{5}}=\dfrac{5}{3}. Thus, the correct answer is D.
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