Step-by-Step Solution
Key Concept: Finding the intersection of two lines by solving their equations simultaneously to locate point $D$.
Given the triangle with vertices $B(3,\sqrt{3})$, $C(0,0)$, and point $A$ above, we find $D$ as the intersection of line $BD: x - \sqrt{3}y = 0$ and line $\sqrt{3}x + y = 2\sqrt{3}$. Solving simultaneously gives $D = \left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)$. The distance $AC = BC = 2\sqrt{3}$ establishes the triangle's properties.
Correct Answer: 2