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AMC 10 (2013) Solutions

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

(\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }56\qquad\textbf{(D) }60\qquad\textbf{(E) }72\qquad )

http://cache.artofproblemsolving.com/asyforum/c/e/e/cee0c4038bfb9b14cfc95050e9b3d20bf808167d.png

Solution: Perimeter = 2(AD + AF). But AD = EF (since ABCD is a parallelogram).
Hence perimeter = 2(AF + EF).
Now ABC is isosceles (AB = AC = 28). Thus angle B = angle C. But EF is parallel to AB. Thus angle FEC = angle B which in turn is equal to angle C.
Hence triangle CEF is isosceles. Thus EF = CF.
Perimeter = 2(AF + EF) = 2(AF + EF) =2AC = (2 \times 28) = 56.

Ans. (C) 56

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