Assume WLOG that $ \angle {B} \leq \angle{C}$ (in this case $ F$ will be on the segment $ AE$, as in the statement). Denote by $ Y$, $ Z$ the tangency points of the incircle with the sides $ CA$, and $ AB$. Let $ T$ be the intersection point of the lines $ YZ$ and $ BC$. Since $ DZEY$ is an harmonic quadrilateral (because of the concurrency of the tangects at $ Y$, $ Z$ on the line $ DE$), the line $ TD$ is tangent to the incircle at $ D$, and thus $ \angle{TED} = \angle{TDE}$. But $ \angle{TED} = \angle{TFE}$, and so the lines $ TD$ and $ CF$ are parallel. Now, since the quadruple $ (B, E, C, T)$ is harmonic, the pencil $ D(B, E, C, T)$ is harmonic, and by intersecting it with the line $ CF$, we conclude that $ F$ is the midpoint of $ CG$ (according to the parallelism of $ DT$ and $ CF$).

If we denote $ K\in\omega(I)\cap BD$,$ L\in\omega(I)\cap CD,N\in\omega(I)\cap AB,M\in\omega(I)\cap AC$,then problem follows by the fact that the following fours are harmonic:$ D(DMEN)\cap l,D(DNKE)\cap l,D(DMLE)\cap l$,where $ l=CF$.

Let the incircle interescts $ BD$ and $ CD$ at $ K,L$ respectively Then $ \frac{GF}{CF}=\frac{DG \times EK}{DC \times EL}=\frac{DL \times EK}{DK \times EL}$ I think I have proved that $ DL \times EK=DK \times EL$, but the proof is long and complicated. Does anyone have a nice proof for this?

See here: http://www.mathlinks.ro/viewtopic.php?t=2619 (Read especially the 5th post). For some extensions, see also here: http://www.mathlinks.ro/viewtopic.php?p=887161#887161.

Let tangent to incircle in a point $ D$ intersect line $ BC$ in $ K$. $ DK\parallel CF$. Line parallel to $ DK$ through $ B$ intersect line $ AE$ in $ M$ Let $ L$ be a 'point' respective to a set of lines parallel to $ DK$. Then $ (GFCL)=(DFEM)=(KCEB)=-1$ So $ GF=FC$

I have made one unnecessary turn. Let tangent to incircle in a point $ D$ intersect line $ BC$ in $ K$. $ DK\parallel CF$. Let $ L$ be a 'point' respective to a set of lines parallel to DK. Then $ (GFCL)=(KCEB)=-1$. So $ GF=FC$.

I will use pohoata's notation. Since $ T$ lies on polar of $ A$, $ A$ lies on polar of $ T$. But $ TE$ is tangent to given circle so $ AE$ is polar of $ T$ and we already have that $ D$ is on both $ AE$ and incircle, so $ TD$ is tangent to incircle. Now it is easy to conclude that $ CF$ and $ TD$ are parallel and by Menelaus theorem: $ \frac{GF}{FC}=\frac{EB}{EC}\frac{DG}{BD}=\frac{BZ}{ZY}\frac{CT}{TB}=\frac{BZ}{TB}\frac{CT}{CY}$ $ =\frac{\sin \angle BTZ}{\sin \angle AZY}\frac{\sin \angle AYZ}{\sin \angle CTY}=1$ which implies $ CF=FG$.

See from the post #2 that $DT\parallel CF$ and $T$ is the harmonic conjugate of $E$ w.r.t. $BC$, then apply Menelaos to $\triangle BCG$ and the transversal $EFD$, tol get $\frac{CE}{BE}\cdot \frac{BD}{DG}\cdot \frac{GF}{FC}=1$, but $\frac{BD}{DG}=\frac{BT}{CT}$ from parallel lines and $\frac{BT}{CT}=\frac{BE}{CE}$ from the harmonic division and remains $\frac{GF}{FC}=1$. Best regards, sunken rock

I will use pohoatza's notation. In addition, let $BD$ intersect the incircle again at $Q$. Let $I$ be the incenter. We have $\angle EQD = \angle CEF = \angle CFE = \angle DFG$, thus, $GQFE$ is cyclic. Furthermore, note that $C$ is the circumcenter of $\triangle EFY$, so $\angle EFY = 180 - \frac {C} {2} $, and $\angle DFY = \frac {C} {2}$. Furthermore, $\angle FDY = 90 - \frac {C} {2}$, so $\angle DYF = 90$. Thus, $\triangle DYE \sim \triangle IYC$. Now I will calculate $GF$. We have $\frac {GF} {QE} = \frac {DF} {DQ}$ from $GQFE$ cyclic, so $GF = QE \cdot \frac {DF} {DQ}$. Now using $DF = \frac {DY} {\sin {\frac {C} {2}}}$, we get $GF = \frac {QE} {DQ} \cdot \frac {DY} {\sin {\frac {C} {2}}}$. $ZE$ is the symmedian of $QEDZ$, so $\frac {QE} {DQ} = \frac {ZE} {2ZD}$. But we also have $\frac {ZE} {ZD} = \frac {AE} {AZ} = \frac {AE} {AY} = \frac {YE} {DY}$. Thus, we can multiply out to get $GF = \frac {YE} {2 \cdot \sin {\frac {C} {2}}} = CE$, as desired.

I found a solution without using harmonic properties, though admittedly quite a bit more computational.

Let the ray $DN$ intersect the circle $\omega$ with center in point $C$ and has radius $CE$, the second time in point $P$. 1)$\angle CNP=\angle DNA=\angle DEN=\frac{\angle FCN}{2} $ so $\angle FNP=\angle FNC+\angle CNP=90^{\circ}-\frac{\angle FCN}{2}+\frac{\angle FCN}{2}= 90^{\circ}$ because $EC=FC=NC$. And since point $C$ is the center of $\omega$, we must have points $F,C,P$ collinear. 2)$\frac{2FC}{FD}=\frac{FP}{FD}=\frac{\sin\angle FDP}{\sin \angle FPD}=\frac{\sin\angle FDP}{\sin \angle DEN}=\frac{EN}{DN}$ 3) Let the incircle touch $AB$ and $AC$ at points $Z,N$ respectively. Let point $M$ be the middle of $ZE$. Then since $BD$ is the symmedian of $\angle ZDE$, we have $\angle ZDM=\angle GDF=90^{\circ}-\angle MDE-\angle \frac{ABC}{2} $.Also we find that $\angle EZD=\angle DNA+\angle ENC=90^{\circ}-\angle \frac{FCE}{2}$. 4)$\angle ZMD =180^{\circ}-\angle ZDM-\angle DZM=\angle \frac{ABC}{2}+\angle \frac{FCE}{2}+\angle MDE$. $ \angle DZA=\angle ZED =\angle ZMD -\angle MDE$. 5) $\angle DGF= \angle FCE+\angle GBC=\angle FCE+\angle ABC-\angle ZBD=\angle FCE+\angle ABC-\angle AZD+\angle ZDB=\angle FCE+\angle ABC-\angle ZMD +\angle MDE+\angle ZDB=\angle \frac{ABC}{2}+\angle \frac{FCE}{2}+\angle ZDB . $. So we found that $\angle DGF=\angle ZMD$ and $\angle ZDM=\angle GDF$ . Which gives us that $\triangle GDF\sim\triangle MDZ$. 6)So $\frac{2GF}{DF}=\frac{2ZM}{ZD}=\frac {ZE}{ZD}=\frac{EN}{DN}=\frac{2FC}{FD}$, because $AE$ is the symmedian of $\angle ZEN$. From (6) we find $GF=FC$ and we are done.

Projective geometry yields the following quick solution. Let the incircle touch $AB$ and $AC$ at $F$ and $G$ respectively. Also, let $BC \cap GF \equiv X$ and $AE \cap GF \equiv Y$. By the Symmedian Lemma, $DFEG$ is harmonic. This implies that $XD=XE$ from tangency, thus $XD$ is parallel to $CF$. Therefore we can write \[ -1 = (X, Y; G, F) = (X, E; C, B) = (\infty, F; C, G) \]and the conclusion follows.

Hmm easy even for a P1 Fang-jh wrote: Let $ ABC$ be a triangle, let $ AB > AC$. Its incircle touches side $ BC$ at point $ E$. Point $ D$ is the second intersection of the incircle with segment $ AE$ (different from $ E$). Point $ F$ (different from $ E$) is taken on segment $ AE$ such that $ CE = CF$. The ray $ CF$ meets $ BD$ at point $ G$. Show that $ CF = FG$. Let the incircle touch $AC,AB$ at $G,H$ respectively.Let $T=GH\cap BC$.Then since $GHDE$ is harmonic hence $TD$ is tangent to the incircle.Hence $TD=TE\implies TD\parallel CF$.Now \[-1=(T,E,B,C)\overset{D}{=}(\infty_{CF},F,G,C) \implies CF=FG\square\]