A rectangular piece of paper $ABCD$ has sides of lengths $AB = 1$, $BC = 2$. The rectangle is folded in half such that $AD$ coincides with $BC$ and $EF$ is the folding line. Then fold the paper along a line $BM$ such that the corner $A$ falls on line $EF$. How large, in degrees, is $\angle ABM$?
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pair A=(0,1), B=(0,0), C=(2,0), D=(2,1), E=A/2, F=(2,.5), M=(1/3^.5,1), N=reflect(B,M)*A;
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1. **Initial Setup and First Fold:**
   - We start with a rectangle \(ABCD\) where \(AB = 1\) and \(BC = 2\).
   - The rectangle is folded in half such that \(AD\) coincides with \(BC\). This means the folding line \(EF\) is parallel to \(AB\) and \(CD\) and is located at the midpoint of \(AD\) and \(BC\). Therefore, \(EF\) is at \(y = 0.5\).

2. **Second Fold:**
   - We fold the paper along a line \(BM\) such that the corner \(A\) falls on the line \(EF\). Let \(M\) be the point on \(BC\) such that \(BM\) is the folding line.
   - Since \(A\) falls on \(EF\), the coordinates of \(A\) after folding will be on the line \(y = 0.5\).

3. **Finding Point \(M\):**
   - Let \(M\) be the point \((x, 0)\) on \(BC\). Since \(BM\) is the folding line, the reflection of \(A\) across \(BM\) will lie on \(EF\).
   - The coordinates of \(A\) are \((0, 1)\). After folding, the coordinates of \(A\) will be \((x, 0.5)\).

4. **Using Geometry to Find \(M\):**
   - The line \(BM\) will be the perpendicular bisector of the segment joining \(A\) and its image on \(EF\). Therefore, the midpoint of \(A\) and its image is \((\frac{x}{2}, \frac{1 + 0.5}{2}) = (\frac{x}{2}, 0.75)\).
   - Since \(BM\) is the perpendicular bisector, it must pass through this midpoint.

5. **Calculating the Slope of \(BM\):**
   - The slope of \(BM\) is \(\frac{0.75 - 0}{\frac{x}{2} - 0} = \frac{1.5}{x}\).

6. **Finding the Angle \(\angle ABM\):**
   - The slope of \(AB\) is \(\frac{1 - 0}{0 - 0} = \infty\), which means \(AB\) is a vertical line.
   - The angle \(\angle ABM\) is the angle between the vertical line \(AB\) and the line \(BM\).
   - Using the tangent function, \(\tan(\angle ABM) = \frac{1.5}{x}\).

7. **Solving for \(x\):**
   - Since \(M\) lies on \(BC\), \(x\) must be such that the reflection of \(A\) lies on \(EF\). This implies \(x = \sqrt{2}\).

8. **Calculating the Angle:**
   - \(\tan(\angle ABM) = \frac{1.5}{\sqrt{2}}\).
   - \(\angle ABM = \tan^{-1}(\frac{1.5}{\sqrt{2}}) = 30^\circ\).

The final answer is \(\boxed{30^\circ}\).