2005 AMC 10B Problem 14

Take $\overline{CD}$ as the base. Since $C$ is the midpoint of $\overline{BD}$ and $BC = 2,$ we have $CD = 2.$ The height of $\triangle CDM$ is the distance from $M$ to line $BD.$ Because $M$ is the midpoint of $\overline{AC},$ this distance is half the height of $\triangle ABC,$ which is $\dfrac12 \cdot \sqrt3 = \dfrac{\sqrt3}{2}.$ The area is $$\dfrac12 \cdot 2 \cdot \dfrac{\sqrt3}{2} = \dfrac{\sqrt3}{2}.$$ Thus, C is the correct answer.