To solve the problem, we can use similar triangles and area calculation:

Step 1: Analyze the square and points

Let the side length of square (ABCD) be (s).

• (E) is on the extension of (CD) such that (DE = CD = s), so (CE = CD + DE = 2s).
• (BE) intersects (AD) at (F).

Step 2: Use similar triangles

Since (AD \parallel BC) (sides of a square), (\triangle DFE \sim \triangle CBE) (common angle at (E), alternate interior angles).
The ratio of similarity is (\frac{DE}{CE} = \frac{s}{2s} = \frac{1}{2}).

Thus, (\frac{FD}{BC} = \frac{1}{2}), so (FD = \frac{BC}{2} = \frac{s}{2}).

Step 3: Calculate area of (\triangle DEF)

Area of (\triangle DEF = \frac{1}{2} \times DE \times FD) (since (FD \perp DE)).
Given area = 1:
[ \frac{1}{2} \times s \times \frac{s}{2} = 1 ]
[ \frac{s^2}{4} = 1 ]
[ s^2 = 4 ]

Answer: (\boxed{4})

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