A rectangle $ABEF$ is drawn on the leg $AB$ of a right triangle $ABC$, whose apex $F$ is on the leg $AC$. Let $X$ be the intersection of the diagonal of the rectangle $AE$ and the hypotenuse $BC$ of the triangle. In what ratio does point $X$ divide the hypotenuse $BC$ if it is known that $| AC | = 3 | AB |$ and $| AF | = 2 | AB |$?
1. **Given Information and Setup:**
   - We have a right triangle \( \triangle ABC \) with \( \angle B = 90^\circ \).
   - \( |AC| = 3|AB| \) and \( |AF| = 2|AB| \).
   - Rectangle \( ABEF \) is drawn such that \( AB \) is one of its sides and \( F \) lies on \( AC \).
   - \( X \) is the intersection of the diagonal \( AE \) of the rectangle and the hypotenuse \( BC \).

2. **Labeling and Coordinates:**
   - Let \( AB = a \). Then \( AC = 3a \) and \( AF = 2a \).
   - Place \( A \) at the origin \((0,0)\), \( B \) at \((a,0)\), and \( C \) at \((0,3a)\).
   - Since \( F \) is on \( AC \) and \( |AF| = 2a \), \( F \) is at \((0,2a)\).

3. **Finding Coordinates of \( E \):**
   - Since \( E \) is directly above \( B \) and forms a rectangle with \( ABEF \), \( E \) is at \((a,2a)\).

4. **Equation of Diagonal \( AE \):**
   - The line \( AE \) passes through points \( A(0,0) \) and \( E(a,2a) \).
   - The slope of \( AE \) is \( \frac{2a - 0}{a - 0} = 2 \).
   - The equation of line \( AE \) is \( y = 2x \).

5. **Equation of Hypotenuse \( BC \):**
   - The line \( BC \) passes through points \( B(a,0) \) and \( C(0,3a) \).
   - The slope of \( BC \) is \( \frac{3a - 0}{0 - a} = -3 \).
   - The equation of line \( BC \) is \( y = -3x + 3a \).

6. **Finding Intersection \( X \):**
   - Set the equations of \( AE \) and \( BC \) equal to find \( X \):
     \[
     2x = -3x + 3a
     \]
     \[
     5x = 3a
     \]
     \[
     x = \frac{3a}{5}
     \]
   - Substitute \( x = \frac{3a}{5} \) back into \( y = 2x \):
     \[
     y = 2 \left( \frac{3a}{5} \right) = \frac{6a}{5}
     \]
   - Thus, \( X \) is at \( \left( \frac{3a}{5}, \frac{6a}{5} \right) \).

7. **Finding the Ratio \( BX : XC \):**
   - The coordinates of \( B \) are \( (a,0) \) and \( C \) are \( (0,3a) \).
   - The distance \( BX \) is:
     \[
     BX = \sqrt{\left( a - \frac{3a}{5} \right)^2 + \left( 0 - \frac{6a}{5} \right)^2} = \sqrt{\left( \frac{2a}{5} \right)^2 + \left( -\frac{6a}{5} \right)^2} = \sqrt{\frac{4a^2}{25} + \frac{36a^2}{25}} = \sqrt{\frac{40a^2}{25}} = \frac{2a\sqrt{10}}{5}
     \]
   - The distance \( XC \) is:
     \[
     XC = \sqrt{\left( 0 - \frac{3a}{5} \right)^2 + \left( 3a - \frac{6a}{5} \right)^2} = \sqrt{\left( -\frac{3a}{5} \right)^2 + \left( \frac{9a}{5} \right)^2} = \sqrt{\frac{9a^2}{25} + \frac{81a^2}{25}} = \sqrt{\frac{90a^2}{25}} = \frac{3a\sqrt{10}}{5}
     \]
   - Therefore, the ratio \( BX : XC \) is:
     \[
     \frac{BX}{XC} = \frac{\frac{2a\sqrt{10}}{5}}{\frac{3a\sqrt{10}}{5}} = \frac{2}{3}
     \]

The final answer is \(\boxed{2:3}\).