
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Let the length of side \( AB \) of parallelogram \( ABCD \) be denoted by \( a \). Let a perpendicular \( BH \) be dropped from vertex \( B \) to line \( CM \). Consider the right triangle \( BHC \).

Given that \( BH = AB = a \). Since \( BC \parallel AB \), then \( \angle BCH = \angle CMD = 30^{\circ} \). Thus, \( BC = 2BH = 2a \). Now consider triangle \( MDC \). Since \( MD = \frac{AD}{2} = \frac{BC}{2} = a \) and \( CD = AB = a \), this triangle is isosceles. Therefore, \( \angle MCD = \angle CMD = 30^{\circ} \). Then

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

and

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

Calculate the area of parallelogram \( ABCD \) given that \( AD = 2 \):

\[ S_{ABCD} = AD \cdot BC \sin 120^{\circ} = 2 \cdot 1 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \]

Present the answer in LaTex format: \boxed{Your answer}
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