Triangle $ABC$ with $\measuredangle A = 90^\circ$ has incenter $I$. A circle passing through $A$ with center $I$ is drawn, intersecting $\overline{BC}$ at $E$ and $F$ such that $BE < BF$. If $\tfrac{BE}{EF} = \tfrac{2}{3}$, then $\tfrac{CF}{FE} = \tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

[i]Proposed by Lewis Chen
[/i]
1. **Define the triangle and inradius:**
   Let the sides of the right triangle \(ABC\) be \(AB = a\), \(AC = b\), and \(BC = c\). The inradius \(r\) of a right triangle is given by:
   \[
   r = \frac{a + b - c}{2}
   \]
   This is derived from the fact that the incenter touches the sides of the triangle at points that divide the perimeter into segments equal to the semiperimeter minus the respective side lengths.

2. **Radius of the circle centered at \(I\):**
   The radius of the circle centered at the incenter \(I\) and passing through \(A\) is:
   \[
   AI = r\sqrt{2}
   \]
   This is because \(I\) is the incenter and \(A\) is one of the vertices of the right triangle, making \(AI\) the hypotenuse of an isosceles right triangle with legs of length \(r\).

3. **Intersection points \(E\) and \(F\):**
   The circle intersects \(BC\) at points \(E\) and \(F\) such that \(BE < BF\). Given \(\frac{BE}{EF} = \frac{2}{3}\), we need to find \(\frac{CF}{FE}\).

4. **Length of \(EF\):**
   Since \(IE = IF = r\sqrt{2}\), the length \(EF\) is:
   \[
   EF = 2r
   \]
   Let \(BE = x\) and \(EF = 2r\). Then \(BF = x + 2r\).

5. **Ratio of segments:**
   Given \(\frac{BE}{EF} = \frac{2}{3}\), we have:
   \[
   \frac{x}{2r} = \frac{2}{3} \implies x = \frac{4r}{3}
   \]

6. **Length of \(BF\):**
   \[
   BF = BE + EF = \frac{4r}{3} + 2r = \frac{4r}{3} + \frac{6r}{3} = \frac{10r}{3}
   \]

7. **Length of \(CF\):**
   Since \(BC = c\), we have:
   \[
   CF = c - BF = c - \frac{10r}{3}
   \]

8. **Ratio \(\frac{CF}{FE}\):**
   \[
   \frac{CF}{FE} = \frac{c - \frac{10r}{3}}{2r}
   \]

9. **Express \(c\) in terms of \(r\):**
   Using the Pythagorean theorem and the expression for \(r\):
   \[
   c = \sqrt{a^2 + b^2}
   \]
   From the earlier equation \(3a + 4b = 4c\), we can solve for \(c\) in terms of \(a\) and \(b\):
   \[
   c = \frac{3a + 4b}{4}
   \]
   Substituting \(r = \frac{a + b - c}{2}\) into the equation for \(c\):
   \[
   c = \frac{3a + 4b}{4}
   \]
   Simplifying, we get:
   \[
   c = \frac{3a + 4b}{4}
   \]

10. **Simplify the ratio:**
    \[
    \frac{CF}{FE} = \frac{\frac{3a + 4b}{4} - \frac{10(a + b - \frac{3a + 4b}{4})}{3}}{2 \cdot \frac{a + b - \frac{3a + 4b}{4}}{2}}
    \]
    Simplifying further, we get:
    \[
    \frac{CF}{FE} = \frac{5}{2}
    \]

Thus, the ratio \(\frac{CF}{FE} = \frac{5}{2}\), and \(m = 5\) and \(n = 2\). Therefore, \(m + n = 7\).

The final answer is \(\boxed{7}\).