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[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

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# Your action step 2

# Your action step 3

...

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Given a trapezoid with labeled points as shown in the diagram, the point of intersection of the extended non-parallel sides is labeled as \(E\), and the point of intersection of the diagonals is labeled as \(F\). Similar right triangles \(BFC\) and \(DFA\) have corresponding sides \(x, y\) in the first triangle and \(4x, 4y\) in the second triangle.

The height of the trapezoid \(h\) is given by the sum of the heights of triangles \(BFC\) and \(AFD\):

\[ h = x y + \frac{4x \cdot 4y}{4} = 5xy \]

The area of the trapezoid is \(\frac{15}{16}\) of the area of triangle \(AED\):

\[ \frac{1}{2} AC \cdot BD = \frac{15}{16} \cdot \frac{1}{2} \cdot AE \cdot ED \cdot \sin 60^\circ = \frac{15 \sqrt{3}}{64} \cdot \frac{4}{3} AB \cdot \frac{4}{3} CD = \frac{15 \sqrt{3}}{36} AB \cdot CD \]

From this, we have:

\[ \frac{25}{2} xy = \frac{15 \sqrt{3}}{36} \sqrt{x^2 + 16y^2} \cdot \sqrt{y^2 + 16x^2} \]

Given that \(x^2 + y^2 = 1\), find:

\[ \frac{30}{\sqrt{3}} xy = \sqrt{1 + 15y^2} \cdot \sqrt{1 + 15x^2} \]

By simplifying:

\[ 300 x^2 y^2 = 1 + 15(x^2 + y^2) + 225 x^2 y^2 \]
\[ 75 x^2 y^2 = 16 \]
\[ 5xy = \frac{4}{\sqrt{3}} \]

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