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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?

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

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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