How prove find this value $|AD|+|DF|+|FA|=2$ Question:

if $ADB$ and $ACE$ are straight lines with $D,E$ and $B,C$ intersecting at $F$.
if $$|AB|=|AC|=1,|AD|+|DE|+|EA|=4$$
show that:
$$|AD|+|DF|+|FA|=2$$


I have read this famous Urquhart's Theorem,say this if
$$|AD|+|DF|=|AC|+|CF|\Longrightarrow |AB|+|BF|=|AE|+|EF|$$
but I use this result also can't solve this problem.and you can fell this problem is very interesting? Following my some idea,
since
$$|AD|+|DE|+|EA|=|AD|+|DF|+|FE|+|AE|=4$$
so
$$\Longrightarrow |AD|+|DF|=4-|AE|-|EF|$$
Note $$|AC|+|CF|=1+|CF|$$
If we want use Urquhart's Theorem,then we must prove following is right
$$4-|AE|-|EF|=1+|CF|\Longleftrightarrow |AE|+|EF|+|CF|=3\Longleftrightarrow |CE|+|EF|+|CF|=2$$
But I fell use this condition can't prove it
 A: Let's add the excircle of $\triangle ADE$ to the picture. $X, Y$ and $G$ are the tangent points.

$AX$ is equal to semiperimeter of $\triangle ADE$ so $AX = AY = 2$.
Moreover, $AB = AC = 1$ so $B$ and $C$ are the middle points of $AX$ and $AY$ respectively. Hence $BC$ is radical axis of that excircle and point $A$. So $FA = FG$.
Finally, your sum becomes $AD + DF + \color{green}{FA} = AD + DF + \color{green}{FG} = AD + DG = AD + DX = AX = 2$.
A: You can use complex numbers to solve this problem. The computation is a little bit complicated. Without loss of generality, let
$$ A=0, B=1, C=e^{i\theta}, D=d, E=ae^{i\theta}. $$
Here $\theta>0, 0<d<1, a>1$. Sine $F$ is the intersection of the segments $BC$ and $DE$, we have
$$ F=(1-s)B+sC=(1-t)D+tE. $$
Separating the real part from the imaginary part, we have
$$ (1-s)+s\cos\theta=(1-t)d+ta\cos\theta, s\sin\theta=ta\sin\theta $$
from which we obtain
$$ t=\frac{1-d}{a-d}, s=\frac{a(1-d)}{a-d}. $$
Hence
$$ F=\frac{(a-1) d+a (1-d) \cos\theta}{a-d}+i\frac{a (1-d) \sin\theta}{a-d}. $$
From $|AD|+|DE|+|EA|=4$, we have
$$ a+d+\sqrt{a^2-2ad\cos\theta+d^2}=4. $$
Thus we obtain
$$ a=\frac{4(2-d)}{4-d-d\cos\theta}. \tag{1}$$
So
\begin{eqnarray}
&&|AD|+|DF|+|FA|\\
&=&d+|F|+|D-F|\\
&=&d+\frac{(1-d)\sqrt{a^2-2ad\cos\theta+d^2}}{a-d}+\frac{\sqrt{d^2-2ad+a^2(1-2d+2d^2)+2(1-a)a(d-1)d\cos\theta}}{a-d}.\tag{2}
\end{eqnarray}
Now using (1) to replace $a$ in (2), we obtain
$$ |AD|+|DF|+|FA|=2. $$
