2010 AMC 10A Problem 14


Triangle ABCABC has AB=2ACAB=2 \cdot AC . Let DD and EE be on AB\overline{AB} and BC\overline{BC} , respectively, such that BAE=ACD\angle BAE = \angle ACD . Let FF be the intersection of segments AEAE and CDCD , and suppose that CFE\triangle CFE is equilateral. What is ACB\angle ACB ?

(A) 60(B) 75(C) 90(D) 105(E) 120\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ


Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.

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Category: AMC 10A
Points: 3
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