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2013 AMC 10A Problem 12

Problem 12 of 25IntermediateGeometry

In ABC,\triangle ABC, AB=AC=28AB=AC=28 and BC=20.BC=20. Points D,D, E,E, and FF are on sides AB,\overline{AB}, BC,\overline{BC}, and AC,\overline{AC}, respectively, such that DE\overline{DE} and EF\overline{EF} are parallel to AC\overline{AC} and AB,\overline{AB}, respectively. What is the perimeter of parallelogram ADEF?ADEF?

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Solution

Note that DBEABC\triangle DBE \sim \triangle ABC and FECABC\triangle FEC \sim \triangle ABC due to the parallel lines. This tells us that DB=DEDB = DE and FE=FC.FE = FC. We have that the perimeter of ADEFADEF is AD+DE+EF+AF AD + DE + EF + AF =AD+DB+FC+AF= AD + DB + FC + AF =AB+AC = AB + AC =56.= 56. Thus, C is the correct answer.

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Concepts: similarity · parallelogram · perimeter

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.