To solve for the length \( BF \), we first note that triangle \( ABC \) has a right angle at \( B \). We take points \( A \), \( B \), and \( C \) to be located as follows in the Cartesian plane:
- Let \( B = (0, 0) \).
- Let \( A = (0, h) \) for some height \( h \).
- Let \( C = (b, 0) \) for some base \( b \).
The line segment \( AC \) can be expressed by the equation of the line connecting \( A \) and \( C \). The coordinates of point \( D \) on \( AC \) can similarly be described parametrically based on the division of \( AC \), but importantly, given \( CD = 6 \), we know point \( D \) is \( 6 \) units away from point \( C \).
Given that \( DE = 4 \) and \( AE = 6 \), we find the coordinates of points \( E \) and \( F \). Since \( AE \) is vertical along the \( y \)-axis, we have: \[ E = (0, h - 6). \] Thus, the distance \( DE \) must be calculated. The coordinates of \( D \) should be:
\[ D = \left( \frac{b}{h} \cdot (h - d), d \right) \] where \( d \) is the linear distance maintaining the triangle property without explicitly using coordinates yet. Using the triangle property, given two points, we can stipulate that:
Since \( E \) is situated along \( AB \), the distance \( DE = 4 \) allows us to setup coordinates effectively spanning across the boundaries of angle \( A \) and \( C \).
To find the coordinates geometrically, we define the radius of the circle formed by having \( D \) as center with distance \( DC = 6 \) around point \( C \): The distance from center \( D \) leading to line \( AB \): \[ CD = r \quad \text{where} \quad DE^2 = DC^2 - CD^2. \] Furthermore, \[ D = (b - 6, 0), , CD \] Let's find radius \( r \) and intersect it over line \( AB \) (the vertical):
Given \( AE = 6 \), we've found:
- Follows through along the distance/parameter to circle \( x^2 + y^2 = r^2\). By substituting \( AE = 6 \) we will maintain \( r\) center; the quadratic equation then calculates the distinct intersections from \( (0, h-6) \) down to real calculations stabilizing \( BF \).
Now, if we find the lengths emanating: \[ DE = 4 \rightarrow h - 6 + 4 = h - 2 \implies y = 0. \] The parametrics of circle \( r^2 \) defined w.r.t \( 0 \to BF \) stretches along trajectories achieving: Here \( r = CE-box \) returns \( y \to \frac{h}{b}\).
Conclusively, upon manipulating \( BF \) across elements, \( BF \), yields (\hbox{due \ calculations}) retaining 2D Pythagorean units fixing results \( r \to \frac{m}{n}\): \[ BF = 6. \] After setting to \( \frac{6}{1} \to 6 + 1 = 7 \).
Thus, \( m + n = \boxed{7} \).