$18 \cdot 86$ In an equilateral $\triangle A B E$, vertex $E$ is inside the square $A B C D$, and $F$ is the intersection of diagonal $B D$ and segment $A E$. If the length of $A B$ is $\sqrt{1+\sqrt{3}}$, then the area of $\triangle A B F$ is
(A) 1 .
(B) $\frac{\sqrt{2}}{2}$.
(C) $\frac{\sqrt{3}}{2}$.
(D) $4-2 \sqrt{3}$.
(E) $\frac{1}{2}+\frac{\sqrt{3}}{4}$.
(29th American High School Mathematics Examination, 1978)
[Solution] Draw the altitude $F G$ of $\triangle A F B$, and let $x$ represent the length of $A G$.
From the right figure, we can see that $F G=\sqrt{3} x$,
and $\sqrt{1+\sqrt{3}}=A B=x(1+\sqrt{3})$,
which means $1+\sqrt{3}=x^{2}(1+\sqrt{3})^{2}$,
dividing both sides by $1+\sqrt{3}$ gives $1=x^{2}(1+\sqrt{3})$,
so
$$
\begin{aligned}
S_{\triangle A B F} & =\frac{1}{2} \cdot A B \cdot F G \\
& =\frac{1}{2} \cdot x^{2}(1+\sqrt{3}) \cdot \sqrt{3} \\
& =\frac{1}{2} \sqrt{3} .
\end{aligned}
$$

Therefore, the answer is $(C)$.