Tough Perimeter

let, $AD=x$ and $FC=y$
and we know $\angle ABC=90^{\circ}$
we can write this equation(using Pythagoras): $x+1)^2 + (y+1)^2=16$ .......... $(1)$
again $\triangle ADE$ and $\triangle EFC$ is similar.
So,using similarity we acn say $\frac{1}{y}= \frac{x}{1}$ or $xy=1$ ...... $(2)$
Now we have to solve $(1)$ and $(2)$ equation.
from equation $1$ ,
$x^2+2x+1+y^2+2y+2 = 16$
or, $x^2+y^2+2(x+y)+2=16$
or, $(x+y)^2-2xy+2(x+y)+2=16$
or, $(x+y)^2-2*1+2(x+y)+2=16$ [using equation $(2)$ ]
or, $(x+y)^2-2+2(x+y)+2=16$
or, $(x+y)^2+2(x+y)-16=0$
or, $a^2+2a-16=0$ [ Let $a=x+y$ ]
so, $a=\sqrt{17}-1 or,a=-\sqrt{17}-1$ [ we can't take this value as it is negative]
so , $x+y=\sqrt{17}-1$
hence,perimeter of this triangle is $\triangle ABC$
$=x+y+DB+BF+AC = \sqrt{17}-1+1+1+4=\sqrt{17} +5$
here $m=17$ and $n=5$
finally, $m+n=17+5=\boxed{22}$