Application of Ceva's Theorem in a triangle Let A,B,C be a triangle and F,G,H the intersection points of the incircle with the sides BC,AC and AB. Now the line G(A,F) is mirrored on the angle bisector in A, such that the line G(A,F') is created. Same is done with G(B,G) (mirrored on angle bisector in B) and G(C,H) (mirrored on angle bisector in C).
Now I want to show that mirrored lines G(A,F'), G(B,G') and G(C,H') intersect in one point.
I know that the angle bisectors intersect in the mid-point of the incircle and I know how to show that G(A,F),G(B,G) and G(C,H) intersect in one point by Ceva Theorem. But at this point I'm stuck.
Can someone help me out? :)
 A: The general result is - as in the quick answer of user above, going straightforward to the point - related to constructing the isogonal conjugate of a point. In this answer, i will go in some detail, giving a solution specific to the special situation we have, and identify the points with those in the Encyclopedia of Triangle Centers, ETC. There, $X(k)$ is a "triangle center", the $k$.th in human list of relevant points in a triangle.

We start with a general triangle $\Delta ABC$ with sides $a,b,c$, and let $s=(a+b+c)/2$ be its semiperimeter.
I will use slightly different notations, $D,E,F$, for the vertices of the intouch (or contact) triangle. (Since $G,H$ are elsewhere on my hard disk.) The picture is as follows:

So $I=X(1)$ is the incenter, $\Delta DEF$ is its pedal triangle, we draw the cevians $AD$, $BE$, $CF$, and they are intersecting in a point, the
Gergonne point
(see also here), listed as $X(7)$ in the ETC. To show it is well defined, i.e. the three above cevians intersect indeed in a point, note that
we have $AE=AF=x$, $BF=BD=y$, $CD=CE=z$ from the properties of tangents from a point to a circle, (here the incircle,) and the theorem of Ceva shows their concurrency, the relation in it being
mot-a-mot
equivalent to
$$
\left(-\frac yz\right) \left(-\frac zx\right) \left(-\frac xy\right)=-1
\ ,
$$
which is true.
We may need something more, so note that solving the system
$y+z=a$, $z+x=b$, $x+y=c$, we obtain explicitly $x=s-a$, $y=s-b$, $z=s-c$.
So the barycentric coordinates of the Gergonne point $X(7)$ are
$$
\left[\ \frac 1{s-a}\ :\ \frac 1{s-b}\ :\ \frac 1{s-c}\right]
=
\left[\ (s-b)(s-c)\ :\ (s-a)(s-c)\ :\ (s-a)(s-b)\ \right]
\ ,
$$
because of
$
\displaystyle
D=
[0:s-c:s-b]=
\left[0:\frac 1{s-b}:\frac 1{s-c}\right]
$, and of the other two similar relations.
Now let us reflect $AD$ w.r.t. the angle bisector $AI$, obtaining a point $D'\in BC$. Similarly construct $E'$, $F'$. Observe that the two angles

*

*$\widehat{BAD}$, and $\widehat{DAC}$ are reflected into

*$\widehat{D'AC}$, and respectively $\widehat{BAD'}$
so the corresponding measures are equal. This gives:
$$
\begin{aligned}
\frac{D'B}{D'C}
&=
\frac{D'B}{AD'}
\cdot
\frac{AD'}{D'C}
=
\frac{\sin\widehat{BAD'}}{\sin B}
\cdot
\frac{\sin C}{\sin \widehat{D'AC}}
=
\frac{\sin\widehat{DAC}}{\sin C}
\cdot
\frac{\sin B}{\sin \widehat{BAD}}
\cdot
\left(\frac{\sin C}{\sin B}\right)^2
\\
&=
\frac{DC}{AD}\cdot\frac{AD}{BD}
\cdot\left(\frac{AB}{AC}\right)^2
=
\frac{s-c}{s-b}\cdot\frac{c^2}{b^2}\ .
\end{aligned}
$$
The other two similar relations for
$\displaystyle\frac{E'C}{E'A}$ and
$\displaystyle\frac{F'A}{F'B}$ are obtained
by cyclic permutation of $a,b,c$ in the formula, and we can apply Ceva
to conclude that $AD'$, $BE'$, $CF'$ intersect in a point.
This is the point $X(55)$ in the ETC.
$\square$

Bonus:
By the above, we also have the barycentric coordinates of $X(55)$.
From the above computation, $D'=[\ 0\ :\ D'C\ :\ D'B\ ]=[\ 0 \ :\ b^2(s-b)\ :\ c^2(s-c)\ ]$, and using also the similar relations for
$E'$, $F'$, we get
$$
X(55) = [\ a^2(s-a) \ :\ b^2(s-b)\ :\ c^2(s-c)\ ]
\ .
$$
In particular, it is now easy to show (analytically) that $X(55)$,
the incenter
$I=X(1)=[a:b:c]$, the circumcenter $O=X(3)=[\sin 2A:\sin 2B:\sin 2C]
=[ \ 2\sin A\cos A\ :\ \dots\ ]=\left[\ a\frac{b^2+c^2-a^2}{bc}\ :\ \dots\ \right]=[\ a^2(b^2+c^2-a^2)\ :\ b^2(a^2+c^2-b^2)\ :\ c^2(a^2+b^2-c^2)\ ]$
are on a line, by checking the vanishing of the corresponding determinant:
$$
0=
\begin{vmatrix}
a & b & c\\
a^2(b+c-a) & b^2(c+a-b) & c^2(a+b-c)\\
a^2(b^2+c^2-a^2) &  b^2(a^2+c^2-b^2) & c^2(a^2+b^2-c^2)
\end{vmatrix}
\ .
$$
To see this, multiply the second line with $(a+b+c)\ne 0$, then subtract it from the third to get a linear independence with the first line.
It turns out that $X(55)$ lies on $IO$ so that its distances to $I$ and $O$ are in the proportion $r:R$, so it is a homothety center for the two circles.
A: Hint:
First apply the usual form of Ceva's theorem to prove that the lines $AF,BG,CH$ are concurrent. Let the concurrence point be $M$.
What you do further is the construction of the isogonal conjugate of $M$. The existence of isogonal conjugate can be proved using the trigonometric form of the Ceva's theorem.
