Solution:

Write $c = AB = 360,$ $a = BC = 507,$ $b = CA = 780.$ Extend $\overline{FD}$ beyond $D$ to meet ray $BA$ beyond $A$ at $G.$ In triangle $BGF,$ segment $\overline{BD}$ is both an angle bisector and an altitude, so $BG = BF = t.$ The bisector also gives $\frac{CD}{DA} = \frac{a}{c},$ so Menelaus' theorem for line $GDF$ crossing triangle $ABC$ says $$\frac{AG}{GB} \cdot \frac{BF}{FC} \cdot \frac{CD}{DA} = \frac{t - c}{t} \cdot \frac{t}{a - t} \cdot \frac{a}{c} = 1, \qquad \text{so} \qquad t = \frac{2ac}{a + c}.$$

Now let $F_1$ be the point on $\overline{AC}$ with $\overline{FF_1} \parallel \overline{BM}.$ Since $E$ lies on $\overline{BM},$ we have $\overline{EM} \parallel \overline{FF_1},$ so triangles $DEM$ and $DFF_1$ are similar and $\frac{DE}{EF} = \frac{DM}{MF_1}.$ The bisector ratio gives $AD = \frac{bc}{a + c},$ so $DM = \frac{b}{2} - \frac{bc}{a+c} = \frac{b(a - c)}{2(a + c)}.$ Also $CF = a - t = \frac{a(a - c)}{a + c},$ so $CF_1 = CM \cdot \frac{CF}{CB} = \frac{b}{2} \cdot \frac{a - c}{a + c}$ and $MF_1 = \frac{b}{2}\left(1 - \frac{a - c}{a + c}\right) = \frac{bc}{a + c}.$

Therefore $$\frac{DE}{EF} = \frac{DM}{MF_1} = \frac{a - c}{2c} = \frac{147}{720} = \frac{49}{240},$$ and $m + n = 49 + 240 = 289.$