6. Let the length of side $AB$ of parallelogram $ABCD$ be denoted by $a$. Drop a perpendicular $BH$ from vertex $B$ to line $CM$. Consider the right triangle $BHC$. According to the problem, $BH = AB = a$. Since $BC \parallel AB$, we have $\angle BHC = \angle CMD = 30^{\circ}$. Therefore, $BC = 2BH = 2a$. Now consider triangle $MDC$. Since $MD = \frac{AD}{2} = \frac{BC}{2} = a$ and $CD = AB = a$, this triangle is isosceles. Thus, $\angle MCD = \angle CMD = 30^{\circ}$. Then

$$
\angle CDA = 180^{\circ} - 2 \cdot 30^{\circ} = 120^{\circ}
$$

$\mathrm{a}$

$$
\angle BAD = 180^{\circ} - \angle CAD = 180^{\circ} - 120^{\circ} = 60^{\circ}
$$

Calculate the area of parallelogram $ABCD$ given that $AD = 2$:

$$
S_{ABCD} = AD \cdot BC \sin 120^{\circ} = 2 \cdot 1 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}
$$
Answer:

1) $60^{\circ}, 120^{\circ}$; 2) $S_{A B C D}=\sqrt{3}$

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