Perimeters!

In the figure, ray $Bx$ is such that $\angle EBx=20{}^\circ$ and $\angle FBx=25{}^\circ$ . It is easy to see that $\angle CBE=20{}^\circ$ and $\angle FBA=25{}^\circ$ . Let $G$ be the intersection of $Bx$ and $EF$ and let ${E}'$ be the foot of the perpendicular from $E$ on $Bx$ .
Right angled triangles $EBC$ and $EB{E}'$ are congruent, thus $B{E}'=BC=1 \ \ \ \ \ (1)$ Similarly, if ${F}'$ is the foot of the perpendicular from $F$ on $Bx$ , triangles $FBA$ and $FB{F}'$ are congruent, thus $B{F}'=BA=1 \ \ \ \ \ (2)$ By $(1)$ and $(2)$ we conclude that $B{E}'= B{F}'$ This means that points ${E}'$ and ${F}'$ coincide and thus $E$ , ${E}'$ , ${F}'$ and $F$ are collinear. Hence, $B{E}'=B{F}'=BG$ .
So, we have two pairs of congruent triangles: $\triangle BCE\cong \triangle BGE \ \ \ \ \ \text{and} \ \ \ \ \ \triangle BAF\cong \triangle BGF$ Consequently, $CE=GE \ \ \ \ \ \text{and} \ \ \ \ \ \ FA=FG$ Now, we tackle the perimeter $P$ of $\triangle FED$ : $\begin{aligned} P & =DF+FE+ED \\ & =DF+FG+GE+ED \\ & =DF+FA+CE+ED \\ & =DA+CD \\ & =\boxed{2} \\ \end{aligned}$

Using the diagram above $y = \tan(25^{\circ})$ and $z = \tan(20^{\circ})$ and

$y = \tan(25^{\circ}) = \tan(45^{\circ} - 20^{\circ}) = \dfrac{1 - \tan(20^{\circ})}{1 + \tan(20^{\circ})} \implies$

$\overline{FD} = 1 - y = \dfrac{1\tan(20^{\circ})}{1 + \tan(20^{\circ})}$ and $\overline{DE} = 1 - \tan(20^{\circ}) \implies$

$\overline{FE} = \dfrac{1 + \tan^2(20^{\circ})}{1 + \tan(20^{\circ})} \implies$

The perimeter $P_{FED} = \dfrac{1 - \tan^2(20^{\circ}) + 2\tan(20^{\circ}) + 1 + \tan^2(20^{\circ})}{1 + \tan(20^{\circ})} =$

$\dfrac{2(1 + \tan(20^{\circ}))}{1 + \tan(20^{\circ})} = \boxed{2}$ .

Let $ED=a = 1 - \tan 20^\circ$ and $FD=b=1-\tan 25^\circ$ . Then the perimeter is:

$\begin{aligned} p & = ED + FD + EF \\ & = ED + FD + \sqrt{ED^2+FD^2} \\ & = a + b + \sqrt{a^2+b^2} \\ & = 1 - \tan 20^\circ + 1 - \tan 25^\circ + \sqrt{(1-tan \20^\circ)^2+(1-\tan 25^\circ)^2} \\ & = 1 - \tan 20^\circ + 1 - \tan (45^\circ - 20^\circ) + \sqrt{(1-tan 20^\circ)^2+(1-\tan (45^\circ - 20^\circ))^2} & \small \blue{\text{Let }t = \tan 20^\circ} \\ & = 1 - t + 1 - \frac {1-t}{1+t} + \sqrt{(1-t)^2 + \left(1-\frac {1-t}{1+t}\right)^2} \\ & = 1 - t + \frac {2t}{1+t} + \sqrt{(1-t)^2 + \left(\frac {2t}{1+t}\right)^2} \\ & = \frac {1-t^2+2t}{1+t} + \sqrt{\frac {(1-t)^2(1+t)^2+4t^2}{(1+t)^2}} \\ & = \frac {1+2t-t^2}{1+t} + \frac {\sqrt{(1-t^2)^2+4t^2}}{1+t} \\ & = \frac {1+2t-t^2}{1+t} + \frac {\sqrt{t^4+2t^2+1}}{1+t} \\ & = \frac {1+2t-t^2+t^2 + 1}{1+t} \\ & = \frac {2(1+t)}{1+t} = \boxed 2 \end{aligned}$