2013 AMC 12A Problems/Problem 9

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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2013 AMC 12A Problems/Problem 9