jmc

Let $I$ be the incenter of $△ABC$, so that $BI$ and $CI$ are angle bisectors of $\angle ABC$ and $\angle ACB$ respectively. Then, $\angle BID=\angle CBI=\angle DBI,$ so $△BDI$ is isosceles, and similarly $△CEI$ is isosceles. It follows that $DE=DB+EC$, so the perimeter of $△ADE$ is $AD+AE+DE=AB+AC=43$. Hence, the ratio of the perimeters of $△ADE$ and $△ABC$ is $\frac{43}{63}$, which is the scale factor between the two similar triangles, and thus $DE=\frac{43}{63}\times 20=\frac{860}{63}$. Thus, $m+n=\boxed{923}$.