2013 AMC 12A Problems/Problem 9

Revision as of 04:26, 7 February 2013 by Epicwisdom (talk | contribs) (Replaced bar with overline in LaTeX)

Note that because $\overline{DE}$ and $\overline{EF}$ are parallel to the sides of $\triangle ABC$, the internal triangles $\triangle BDE$ and $\triangle EFC$ are similar to $\triangle ABC$, and are therefore also isosceles triangles.

It follows that $BD = DE$. Thus, $AD + DE = AD + DB = AB = 28$.

Since opposite sides of parallelograms are equal, the perimeter is $2 * (AD + DE) = 56$.

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