2005 AMC 10A Problems/Problem 23

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Problem

$BCDE$ is a square. Point $A$ is chosen outside of $BCDE$ such that angle $BAC= 120^\circ$ and $AB=AC$. Point $F$ is chosen inside $BCDE$ such that the triangles $ABC$ and $FCD$ are congruent. If $AF=20$, compute the area of $BCDE$.

(A)1/6 (B) 1/4 (C) 1/3 (D) 1/2 (E)2/3

$\mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3}$

Solution

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See also