Problem
1978 AHSME Problem 23
![[asy] size(100); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((0,1)--(1,0)); draw((0,0)--(.5,sqrt(3)/2)--(1,0)); label("$A$",(0,0),SW); label("$B$",(1,0),SE); label("$C$",(1,1),NE); label("$D$",(0,1),NW); label("$E$",(.5,sqrt(3)/2),E); label("$F$",intersectionpoint((0,0)--(.5,sqrt(3)/2),(0,1)--(1,0)),2W); //Credit to chezbgone2 for the diagram [/asy]](/static/aops/ce2ed24876d91f9449ac74cf90fe3b9b4487ab1e.png)
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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