Difference between revisions of "1978 AHSME Problems/Problem 23"

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Latest revision as of 16:41, 18 June 2021


Vertex $E$ of equilateral $\triangle ABE$ is in the interior of square $ABCD$, and $F$ is the point of intersection of diagonal $BD$ and line segment $AE$. If length $AB$ is $\sqrt{1+\sqrt{3}}$ then the area of $\triangle ABF$ is

$\textbf{(A) }1\qquad \textbf{(B) }\frac{\sqrt{2}}{2}\qquad \textbf{(C) }\frac{\sqrt{3}}{2}\qquad \textbf{(D) }4-2\sqrt{3}\qquad  \textbf{(E) }\frac{1}{2}+\frac{\sqrt{3}}{4}$


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