Unit square $ABCD$ is divided into four rectangles by $EF$ and $GH$, with $BF = \frac14$. $EF$ is parallel to $AB$ and $GH$ parallel to $BC$. $EF$ and $GH$ meet at point $P$. Suppose $BF + DH = FH$, calculate the nearest integer to the degree of $\angle FAH$.
[asy]
size(100); defaultpen(linewidth(0.7)+fontsize(10)); 
pair D2(pair P) {
dot(P,linewidth(3)); return P;
}
// NOTE: I've tampered with the angles to make the diagram not-to-scale. The correct numbers should be 72 instead of 76, and 45 instead of 55. 
pair A=(0,1), B=(0,0), C=(1,0), D=(1,1), F=intersectionpoints(A--A+2*dir(-76),B--C)[0], H=intersectionpoints(A--A+2*dir(-76+55),D--C)[0], E=F+(0,1), G=H-(1,0), P=intersectionpoints(E--F,G--H)[0];
draw(A--B--C--D--cycle);
draw(F--A--H); draw(E--F); draw(G--H); 
label("$A$",D2(A),NW);
label("$B$",D2(B),SW);
label("$C$",D2(C),SE);
label("$D$",D2(D),NE);
label("$E$",D2(E),plain.N);
label("$F$",D2(F),S);
label("$G$",D2(G),W);
label("$H$",D2(H),plain.E);
label("$P$",D2(P),SE);
[/asy]
1. **Identify the given information and set up the problem:**
   - We have a unit square \(ABCD\).
   - \(EF\) is parallel to \(AB\) and \(GH\) is parallel to \(BC\).
   - \(BF = \frac{1}{4}\).
   - \(BF + DH = FH\).

2. **Determine the coordinates of the points:**
   - Since \(ABCD\) is a unit square, we can place the points as follows:
     \[
     A = (0, 1), \quad B = (0, 0), \quad C = (1, 0), \quad D = (1, 1)
     \]
   - \(F\) is on \(BC\) such that \(BF = \frac{1}{4}\), so \(F = \left(\frac{1}{4}, 0\right)\).
   - \(E\) is directly above \(F\) on \(AD\), so \(E = \left(\frac{1}{4}, 1\right)\).

3. **Determine the coordinates of \(H\):**
   - Let \(H\) be on \(CD\) such that \(DH = x\). Since \(BF + DH = FH\), we have:
     \[
     \frac{1}{4} + x = FH
     \]
   - Since \(H\) is on \(CD\), \(H = (1, 1 - x)\).

4. **Calculate \(FH\):**
   - The distance \(FH\) can be calculated using the distance formula:
     \[
     FH = \sqrt{\left(1 - \frac{1}{4}\right)^2 + (1 - x)^2} = \sqrt{\left(\frac{3}{4}\right)^2 + (1 - x)^2} = \sqrt{\frac{9}{16} + (1 - x)^2}
     \]
   - Given \(BF + DH = FH\), we substitute \(BF = \frac{1}{4}\) and \(DH = x\):
     \[
     \frac{1}{4} + x = \sqrt{\frac{9}{16} + (1 - x)^2}
     \]

5. **Solve for \(x\):**
   - Square both sides to eliminate the square root:
     \[
     \left(\frac{1}{4} + x\right)^2 = \frac{9}{16} + (1 - x)^2
     \]
   - Expand both sides:
     \[
     \left(\frac{1}{4} + x\right)^2 = \frac{1}{16} + \frac{1}{2}x + x^2
     \]
     \[
     \frac{1}{16} + \frac{1}{2}x + x^2 = \frac{9}{16} + 1 - 2x + x^2
     \]
   - Simplify and solve for \(x\):
     \[
     \frac{1}{16} + \frac{1}{2}x = \frac{25}{16} - 2x
     \]
     \[
     \frac{1}{2}x + 2x = \frac{25}{16} - \frac{1}{16}
     \]
     \[
     \frac{5}{2}x = \frac{24}{16}
     \]
     \[
     \frac{5}{2}x = \frac{3}{2}
     \]
     \[
     x = \frac{3}{5}
     \]

6. **Determine the coordinates of \(H\):**
   - Since \(x = \frac{3}{5}\), \(H = (1, 1 - \frac{3}{5}) = (1, \frac{2}{5})\).

7. **Calculate \(\angle FAH\):**
   - Use the slopes of \(AF\) and \(AH\):
     \[
     \text{slope of } AF = \frac{1 - 0}{0 - \frac{1}{4}} = -4
     \]
     \[
     \text{slope of } AH = \frac{1 - \frac{2}{5}}{0 - 1} = -\frac{3}{5}
     \]
   - The angle between two lines with slopes \(m_1\) and \(m_2\) is given by:
     \[
     \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|
     \]
     \[
     \tan \theta = \left| \frac{-4 - \left(-\frac{3}{5}\right)}{1 + (-4) \left(-\frac{3}{5}\right)} \right| = \left| \frac{-4 + \frac{3}{5}}{1 + \frac{12}{5}} \right| = \left| \frac{-\frac{20}{5} + \frac{3}{5}}{1 + \frac{12}{5}} \right| = \left| \frac{-\frac{17}{5}}{\frac{17}{5}} \right| = 1
     \]
   - Therefore, \(\theta = 45^\circ\).

The final answer is \(\boxed{45^\circ}\).