$CD$ and $CE$ are tangents at $(I)\Rightarrow FC$ is symmedian in $\triangle{FED}$ ....

That's not really clear or not really correct!

PP. Let $ABC$ be a triangle with the incircle $w$ what touch the sides $BC$ , $CA$ , $AB$ at $D$ , $E$ , $F$ respectively. Let $G$ be the midpoint of $[DE]$ . Prove that $\widehat{EFC}\equiv\widehat{GFD}$ . Quote: Lemma. Let $ABC$ be a triangle with the circumcircle $w$ . Denote $T\in BB\cap CC$ and $S\in AT\cap BC$ , where $XX$ is the tangent line to the circle $w$ in the point $X\in w$ . Then the ray $[AS$ is a symmedian in the triangle $ABC$ .

Proof of PP. Apply upper lemma to $\triangle DEF$ , where $C\in DD\cap EE$ . Obtain that $CF$ is $F$-symmedian, i.e. $\widehat{EFC}\equiv\widehat{GFD}$ . See PP6 from here.

Draw the line through $C$ antiparallel to $DE$ in $\triangle DFE$, so that when we extend $FE$ and $FD$ to meet it at $H$ and $K$, $\triangle FHK \sim \triangle FDE$. Then $\angle CHE= \angle EDF=\angle FEA= \angle CEH$, so $\triangle CEH$ is isoceles, and similarly $\triangle CDK$ is isoceles. Thus $CH=CE=CD=CK$, so $FG$ and $FC$ are medians of similar triangles and thus $\angle EFC= \angle GFD$.

Did inversion for fun. Better labeling wrote: Given a triangle $ABC$, let its incircle touch the sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Let $G$ be the midpoint of the segment $EF$. Prove that $\angle EDG = \angle ADF$. Let $P$ be the intersection of $BC$ and $EF$. Let $K$ be the intersection of $AD$ and incircle, by La Hire's, $KP$ is tangent to incircle. We perform an inversion at $P$ with radius $PD$. Note that $K$ and $D$ are fixed, $E,F$ are swapped. Also, as $PKGID$ is cyclic, we get that $(PKGID)$ maps to $KD$, hence $G$ maps to $KD\cap EF$. Hence, $$\measuredangle EDG=\measuredangle EDP-\measuredangle GDP\overset{\text{inversion}}{=}\measuredangle PFD-\measuredangle(EF,KD)=\measuredangle KDF,$$we are done. [asy][asy]import olympiad; size(12cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair A,B,C,I,D,E,F,G,P,K; A=dir(120);B=dir(210);C=dir(330);I=incenter(A,B,C);D=foot(I,C,B);E=foot(I,A,C);F=foot(I,A,B);G=midpoint(E--F);P=extension(C,B,E,F);K=intersectionpoints(incircle(A,B,C),A--D)[0]; draw(A--B--C--cycle,deep);draw(incircle(A,B,C),org); draw(B--P--E,deep);draw(G--D,med);draw(E--D,med);draw(D--F,med);draw(anglemark(A,D,F),light);draw(anglemark(E,D,G),light);draw(A--D,med);draw(K--P,deep); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$G$",G,dir(G)); dot("$P$",P,dir(P)); dot("$K$",K,dir(K)); dot("$I$",I,dir(I)); [/asy][/asy]

Probably not gonna get a reply, but whatever. @#3, I think #2 is not ultra clear or formal, but it is indeed correct. Viewing the problem in terms of $\triangle DEF$ reduces it to a well-known lemma.

Also, I have not idea why I worked so hard for this type of problem. The problem is essentially Lemma 4.24 in EGMO. We can see this if we reword the problem as follows: Reworded Problem (EGMO Lemma 4.24) wrote: In $\triangle DEF$, let $C$ be the intersections of the tangents to $(DEF)$ at $D$ and $E$. Let $FG$ be the $F$-median. Show that $FG$ and $FC$ are isogonal (i.e. $FC$ is the $F$-symmedian). Hence, we will just provide proofs to EGMO Lemma 4.24 as the problem follows. We will actually present three proofs, one by trigonometry, one by barycentric coordinates, and one by complex numbers. In the first two solutions, we will solve the problem in term of $\triangle ABC$ (with $X$ as the intersection of the tangents to $(ABC)$ at $B$ and $C$ and $M$ as the midpoint of $\overline{BC}$). In the third solution we will solve the $\triangle DEF$ version of this problem as stated above. EGMO Lemma 4.24 wrote: In $\triangle ABC$, let $X$ be the intersections of the tangents to $(ABC)$ at $B$ and $C$. Let $AM$ be the $A$-median. Show that $AM$ and $AX$ are isogonal (i.e. $AX$ is the $A$-symmedian). Solution 1: Trigonometry. Let $M$ be the intersection of the isogonal line of $AX$ and line $BC$. It suffices to show that $M$ is the midpoint of $\overline{BC}$. In the following computation, all angles are directed. By the Law of Sines, $$\frac{\sin\angle MAC}{MC}=\frac{\sin \angle C}{AM};\frac{\sin\angle BAM}{BM}=\frac{\sin \angle B}{AM}$$so $$\frac{MC}{BM}\cdot \frac{\sin\angle BAM}{\sin\angle MAC}=\frac{\sin \angle B}{\sin \angle C}\iff \frac{MC}{BM}=\frac{\sin \angle B \sin\angle BAX}{\sin\angle C\sin \angle XAC}.$$ By the Law of Sines, $$\frac{\sin\angle XAC}{XC}=\frac{\sin\angle ACX}{XA};\frac{\sin\angle BAX}{BX}=\frac{\sin\angle XBA}{AX}$$and rearranging gives $$\frac{XA}{XC}=\frac{\sin\angle ACX}{\sin\angle XAC};\frac{BX}{AX}=\frac{\sin\angle BAX}{\sin\angle XBA}$$so we can conclude $$\frac{\sin\angle ACX\sin\angle BAX}{\sin\angle XAC\sin\angle XBA}=1.$$ But (key step), note that $\angle ACX=\angle BCX+\angle CBA=\angle A+\angle C=180^\circ-\angle B$ so $\sin\angle ACX=\sin\angle B$. Doing the same thing for $\sin\angle XBA=\sin\angle C$ gives us the desired $$\frac{MC}{BM}=\frac{\sin \angle B \sin\angle BAX}{\sin\angle C\sin \angle XAC}=\frac{\sin\angle ACX\sin\angle BAX}{\sin\angle XAC\sin\angle XBA}=1$$and we are done! $\square$ Solution 2: Barycentric Coordinates. Let $\triangle ABC$ be the reference triangle. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ and let $BC=a,CA=b,AB=c$. By EGMO 7.26, the equation of the tangent to $(ABC)$ at $B$ is $a^2z+c^2x=0$ and the equation of the tangent to $(ABC)$ at $C$ is $a^2y+b^2x=0$. From this, we have $X=(-a^2:b^2:c^2)$. Let $G=(1:1:1)$ be the centroid of $\triangle ABC$. Since $G$ lies on the $A$-median, it suffices to show that its isogonal conjugate, the symmedian point, lies on line $AX$. By EGMO 7.6, the symmedian point, $K$, has coordinates $K=(a^2:b^2:c^2)$. Hence, it suffices to check that $$\begin{vmatrix}1&0&0\\a^2&b^2&c^2\\-a^2&b^2&c^2\end{vmatrix}=0\iff b^2c^2-b^2c^2=0$$which is evidently true. $\square$ Reworded Problem (EGMO Lemma 4.24) wrote: In $\triangle DEF$, let $C$ be the intersections of the tangents to $(DEF)$ at $D$ and $E$. Let $FG$ be the $F$-median. Show that $FG$ and $FC$ are isogonal (i.e. $FC$ is the $F$-symmedian). Solution 3: Complex Numbers. Set $(DEF)$ as the unit circle with $D=d,E=e,F=f$. By the midpoint formula, $G=\frac{d+e}2$, and by EGMO Lemma 6.19, $C=\frac{2de}{d+e}$. We want to show that $\measuredangle EFC=\measuredangle GFD$ which is the same as showing $\text{arg}\left(\frac{f-e}{f-c}\right)=\text{arg}\left(\frac{f-g}{f-d}\right)$ which is the same as showing $\frac{f-e}{f-c}:\frac{f-g}{f-d}\in\mathbb{R}$ which is the same as showing $\frac{(f-e)(f-d)}{(f-c)(f-g)}=\overline{\left(\frac{(f-e)(f-d)}{(f-c)(f-g)}\right)}$ which is the same as showing $\frac{f^2-f(d+e)+de}{f^2-f\left(\frac{d^2+6de+e^2}{2d+2e}\right)+de}=\overline{\left(\frac{f^2-f(d+e)+de}{f^2-f\left(\frac{d^2+6de+e^2}{2d+2e}\right)+de}\right)}$. Observe $$\overline{\left(\frac{f^2-f(d+e)+de}{f^2-f\left(\frac{d^2+6de+e^2}{2d+2e}\right)+de}\right)}= \frac{\frac{1}{f^2}-\frac{1}{fd}-\frac{1}{fe}+\frac{1}{de}}{\frac{1}{f^2}-\frac{1}{f}\left(\frac{\frac{1}{d^2}+\frac{6}{de}+\frac{1}{e^2}}{\frac{2}{d}+\frac{2}{e}}\right)+\frac{1}{de}}.$$ The numerator simplifies to $$\frac{1}{f^2}-\frac{1}{fd}-\frac{1}{fe}+\frac{1}{de}=\frac{d^2e^2f^2-f^3e^2d-f^3ed^2+f^4de}{f^4e^2d^2}=\frac{de-fe-fd+f^2}{f^2de}$$and the denominator simplifies to \begin{align*}\frac{1}{f^2}-\frac{1}{f}\left(\frac{\frac{1}{d^2}+\frac{6}{de}+\frac{1}{e^2}}{\frac{2}{d}+\frac{2}{e}}\right)+\frac{1}{de}&=\frac{1}{f^2}-\frac{1}{f}\left(\frac{\frac{de^3+6d^2e^2+ed^3}{d^3e^3}}{\frac{2e+2d}{de}}\right)+\frac{1}{de}\\&= \frac{1}{f^2}-\frac{\frac{e^2+6de+d^2}{2d+2e}}{def}+\frac{1}{de}\\&= \frac{fe^2d^2-f^2de\left(\frac{e^2+6de+d^2}{2d+2e}\right)+f^3de}{d^3d^2e^2}\\&=\frac{de-f\left(\frac{e^2+6de+d^2}{2d+2e}\right)+f^2}{f^2de}\end{align*}hence, $$\overline{\left(\frac{f^2-f(d+e)+de}{f^2-f\left(\frac{d^2+6de+e^2}{2d+2e}\right)+de}\right)}= \frac{\frac{1}{f^2}-\frac{1}{fd}-\frac{1}{fe}+\frac{1}{de}}{\frac{1}{f^2}-\frac{1}{f}\left(\frac{\frac{1}{d^2}+\frac{6}{de}+\frac{1}{e^2}}{\frac{2}{d}+\frac{2}{e}}\right)+\frac{1}{de}}=\frac{\frac{de-fe-fd+f^2}{f^2de}}{\frac{de-f\left(\frac{e^2+6de+d^2}{2d+2e}\right)+f^2}{f^2de}}=\frac{f^2-f(d+e)+de}{f^2-f\left(\frac{d^2+6de+e^2}{2d+2e}\right)+de}$$and we are finally done!$\square$