Three, (This question 20 points) Given a triangle with one side length of 2, the median on this side is 1, and the sum of the other two sides is $1+\sqrt{3}$. Find the area of the triangle.


E, from $A D = B I = 1 \times = 1$, we have $\angle B = \angle 1, \angle C = \angle 2$.
Thus, $\angle 1 + \angle 2 = 90^{\circ}$.
Also, since $A C + A B$
$$
\begin{array}{l}
=1+\sqrt{3}, \\
\therefore(A C+A B)^{2}=(1+\sqrt{3})^{2} .
\end{array}
$$

That is, $A C^{2}+A B^{2}+2 A C \cdot A B=4+2 \sqrt{3}$.
By the Pythagorean theorem, we have $A C^{2}+A B^{2}=B C^{2}=4$.
Substituting (2) into (1), we get $A C \cdot A B=\sqrt{3}$.
$$
\therefore S_{\triangle A B C}=\frac{1}{2} A B \cdot A C=\frac{\sqrt{3}}{2} \text {. }
$$