smc

Let $AC=x.$ By Angle Bisector Theorem, we have $\frac{AB}{AC}=\frac{BD}{CD},$ from which $BD=CD\cdot \frac{AB}{AC}=\frac{30}{x}.$ Recall that $x>0.$ We apply the Triangle Inequality to $△ABC:$ 1. $AC+BC>AB⟺x+\left(\frac{30}{x}+3\right)>10$ We simplify and complete the square to get ${\left(x-\frac{7}{2}\right)}^2+\frac{71}{4}>0,$ from which $x>0.$ 2. $AB+BC>AC⟺10+\left(\frac{30}{x}+3\right)>x$ We simplify and factor to get $(x+2)(x-15)<0,$ from which $0<x<15.$ 3. $AB+AC>BC⟺10+x>\frac{30}{x}+3$ We simplify and factor to get $(x+10)(x-3)>0,$ from which $x>3.$ Taking the intersection of the solutions gives $$(m,n)=(0,\infty )\cap (0,15)\cap (3,\infty )=(3,15),$$ so the answer is $m+n=\boxed{18}.$