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
$\textbf{(A) }1\qquad \textbf{(B) }\frac{\sqrt{2}}{2}\qquad \textbf{(C) }\frac{\sqrt{3}}{2}\qquad \textbf{(D) }4-2\sqrt{3}\qquad  \textbf{(E) }\frac{1}{2}+\frac{\sqrt{3}}{4}$
Place square ABCD on the coordinate plane with A at the origin. 
In polar form, line BD is rsinθ=sqrt(1+sqrt(3))-rcosθ and line AF is θ=(pi)/3
This means that the length of the intersection (r) is
r*(sqrt3)/2=sqrt(1+sqrt(3))-r/2
Solving for r you get: r=2/sqrt(1+sqrt(3))
Using the firmly for area of a triangle (A=ab(Sinθ)/2) you get A=2/sqrt(1+sqrt(3))*sqrt(1+sqrt(3))*sin(pi/3)/2=(sqrt3)/2
Getting C as the answer