<s>[INST] How to prove that the segment $FG= DJ+EH$ 
$BE, CD, AF$ are the angle bisectors of the $\triangle ABC$. A circle passes through the points $D, E, F$ and intersects the triangle again at $J, H, G$.  How to prove that $FG=DJ+EH$?
I tried to use  angle bisector theorem with  intersecting secant theorem and got some equations. Am I on the right track? Any hints or answers?
 [/INST] 
Let the angle bisectors of $\triangle ABC$ meet the opposite sides at $D$, $E$, $F$, and let $\bigcirc DEF$ meet the sides again at $D'$, $E'$, $F'$. Define signed lengths $d:=|DD'|$, $e:=|EE'|$, $f:=|FF'|$, so that the task is to show $d+e+f=0$.
By the Intersecting Secants Theorem, using signed distances oriented with the vertex order $ABC$, we have
$$\begin{align}
|AF||AF'| = |EA||E'A| &\quad\to\quad |AF|(|AF|+|FF'|)=|EA|(|E'E|+|EA|) \\
&\quad\to\quad |AF|(|AF|+f)=|EA|(-e+|EA|) \tag{1}
\end{align}$$
By the Angle Bisector Theorem,
$$|AF| = |AB|\cdot \frac{|CA|}{|BC|+|CA|}=\frac{bc}{a+b} \qquad\qquad
|EA| = \frac{bc}{c+a} \tag{2}$$
so that $(1)$ becomes
$$(a + b)^2 (a + c) e +(a + b) (a + c)^2 f = b c (b - c) (2 a + b + c) \tag{3}$$
Likewise,
$$\begin{align}
(b + c)^2 (b + a) f +(b + c) (b + a)^2 d &= c a (c - a) (a + 2b + c) \\
(c + a)^2 (c + b) d +(c + a) (c + b)^2 e &= a b (a - b) (a + b + 2c) 
\end{align}\tag{4}$$
Solving yields
$$(d,e,f) = \frac{
\left(\;
 (b - c) (a b c - p^3 + 2 a p^2),
 (c - a) (a b c - p^3 + 2 b p^2),
 (a - b) (a b c - p^3 + 2 c p^2)
 \;\right)}{2 (a + b) (b + c) (c + a)}\tag{5}$$
where $p:=a+b+c$. From here, we readily see that $d+e+f=0$, as desired. $\square$
</s>