Note circumcircle of medial triangle is nine point circle of $\triangle ABC$. $I$ be incenter of $\triangle ABC$. Incircle touch $AB$ at $D$. Let $U=IC \cap AB$ and tangent from $U$ to incircle touch at $G$,$G$ not on $AB$. $G',D'$ be antipode of $G,D$ in incircle respectively. by feuerbach theorem we know nine point circle touch incircle at only one point ($F$).[1] Claim 1: $\overline{O-G-F}$ Let $E'$ we reflection of $D$ above $O$. Now take inversion circle $\omega$ center at $O$ with radius $OD$. From $ID \perp DO$ we get $\omega$ is orthogonal to incircle. If $C'$ is feet of altitude from $C$ to $AB$ then it well know that $(C',U;D,E'=-1)$ hence by harmonic conjugate property [2] $OU.OC'=OD^2$. Let $L$ is another intersection of $OF$ with incircle.As $\omega$ is orthogonal to incircle $OD^2=OL.OF$ but $F$ also lie on nine point circle, Therefore image of nine point circle is line $UL$ under inversion. As nine point circle tangent to incircle at one point we get $UL$ is tangent to incircle which can be $L$ only hence $L=G$ (As $UL$ and $UG$ both tangent to incircle and both are not on $AB$). Its well know $\overline{C-D'-E-E'}$[3] from $\measuredangle{CC'O}=\measuredangle{CFO}=90=\measuredangle{CFG}$ we get $\overline{C-G'-F}$. Note from $UD,UG$ are tangent to incircle $IU \perp DG \Rightarrow IC \perp D'G'$ and its also perpendicular bisector. Hence $CG'=CD'$ $\measuredangle{CG'D'}=\measuredangle{D'G'F}=\measuredangle{G'D'C}=\measuredangle{ED'G'}$ which give us $CG'=CD'$ and $G'F=D'E$. $CF=CE$. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(21.601579787698228cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 2.6262361244059194, xmax = 28.22781591210415, ymin = -7.052208706396606, ymax = 6.324920068929267; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((8.321736682647568,-4.)--(8.321736682647568,-3.6782633173524313)--(8.,-3.6782633173524313)--(8.,-4.)--cycle, linewidth(2.) + qqwuqq); /* draw figures */ draw(circle((10.699264745632277,-1.3007352543677215), 2.699264745632279), linewidth(0.4) + zzttqq); draw(circle((10.75,-2.), 3.400367627183861), linewidth(0.4) + yqqqyq); draw((8.,-4.)--(10.503932666590744,1.3914526190747263), linewidth(0.4) + red); draw((8.,-4.)--(13.391452619074734,-1.4960673334092491), linewidth(0.4) + red); draw((10.503932666590744,1.3914526190747263)--(13.5,0.), linewidth(0.4)); draw((13.5,0.)--(13.391452619074734,-1.4960673334092491), linewidth(0.4)); draw((xmin, 1.*xmin-12.)--(xmax, 1.*xmax-12.), linewidth(0.4)); /* line */ draw((xmin, -1.375*xmin + 18.)--(xmax, -1.375*xmax + 18.), linewidth(0.4) + yqqqyq); /* line */ draw(circle((13.5,0.), 1.5), linewidth(0.4) + qqwuqq); draw((12.882257546512566,0.2868958735452213)--(8.516271944751987,-2.88836638228066), linewidth(0.4)); draw((9.111633617719344,-3.4837280552480117)--(12.286895873545218,0.8822575465125695), linewidth(0.4)); draw((8.516271944751987,-2.88836638228066)--(9.111633617719344,-3.4837280552480117), linewidth(0.4)); draw((12.882257546512566,0.2868958735452213)--(13.391452619074734,-1.4960673334092491), linewidth(0.4)); draw((8.,4.)--(8.,-4.), linewidth(0.4)); draw((8.,-4.)--(19.,-4.), linewidth(0.4)); draw((19.,-4.)--(8.,4.), linewidth(0.4)); /* dots and labels */ dot((8.,4.),dotstyle); label("$A$", (7.75261880701374,4.338067609101955), NE * labelscalefactor); dot((19.,-4.),dotstyle); label("$B$", (19.127728309841746,-4.170514299013481), NE * labelscalefactor); dot((8.,-4.),dotstyle); label("$C$", (7.661617930991116,-4.261515175036107), NE * labelscalefactor); dot((13.5,0.),linewidth(4.pt) + dotstyle); label("$O$", (13.561508059791242,0.12169368672033064), NE * labelscalefactor); dot((10.699264745632277,-1.3007352543677215),linewidth(4.pt) + dotstyle); label("$I$", (10.755647715760334,-1.1826522029372943), NE * labelscalefactor); dot((13.391452619074734,-1.4960673334092491),linewidth(4.pt) + dotstyle); label("$E$", (13.455340371098181,-1.3798207676529817), NE * labelscalefactor); dot((10.503932666590744,1.3914526190747263),linewidth(4.pt) + dotstyle); label("$F$", (10.43714464968115,1.5473740777414555), NE * labelscalefactor); dot((12.286895873545218,0.8822575465125695),linewidth(4.pt) + dotstyle); label("$D$", (11.938659104054446,0.7283661935378306), NE * labelscalefactor); dot((8.,-1.3007351418664217),linewidth(4.pt) + dotstyle); label("$B'$", (8.055955060422487,-1.1826522029372943), NE * labelscalefactor); dot((10.699264868241292,-4.),linewidth(4.pt) + dotstyle); label("$A'$", (10.755647715760334,-3.882344858275169), NE * labelscalefactor); dot((12.631578947368421,0.6315789473684219),linewidth(4.pt) + dotstyle); label("$U$", (12.818334238939812,0.5918648795038931), NE * labelscalefactor); dot((12.882257546512566,0.2868958735452213),linewidth(4.pt) + dotstyle); label("$G$", (12.66666611223544,-0.029974439984044348), NE * labelscalefactor); dot((14.713104126454784,-0.8822575465125695),linewidth(4.pt) + dotstyle); label("$E'$", (14.562517696040107,-0.6669805721424193), NE * labelscalefactor); dot((8.516271944751987,-2.88836638228066),linewidth(4.pt) + dotstyle); label("$G'$", (8.571626691217357,-2.760000720662794), NE * labelscalefactor); dot((9.111633617719344,-3.4837280552480117),linewidth(4.pt) + dotstyle); label("$D'$", (9.178299198034852,-3.366673227480294), NE * labelscalefactor); dot((14.052278372764757,-1.18916243996091),linewidth(4.pt) + dotstyle); label("$J$", (14.213681004620048,-1.0309840762329192), NE * labelscalefactor); dot((11.805405405405391,1.232432432432443),linewidth(4.pt) + dotstyle); label("$C'$", (11.86282504070226,1.350205513025768), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] [1].1:Euclidean Geometry in Mathematical Olympiad,Proposition 6.25 (The Feuerbach Tangency) [2]:Euclidean Geometry in Mathematical Olympiads,Lemma 9.17 (Midpoint Lengths) [3]:Euclidean Geometry in Mathematical Olympiads,Lemma 4.9 (The Diameter of the Incircle)

I think I complex bash way too much

Attachments:
P14.pdf (241kb)

I just wanted to point out that this is essentially ELMO 2016/6 config. Let $D$ is foot of altitude from $BC$ and let $G,H$ be the incenter, excenter of $AOD$ notice $\measuredangle OFA = 90^\circ$ so $AF$ is actually the line joining $C$ and mixti touch point. Done. Remark: Actually , it is well known that in right triangle $CT \cap \omega$ is the feurbach point.

Sol:- It isn't hard to note that $CE$ is $C$ nagel cevian. Since $\angle C=90^\circ$ , homothety at $C$ with ratio $2$ maps nine point circle to circumcircle and incircle to $C$ mixtilinear incircle (since radius of mixtilinear incircle $= \frac{r}{\cos^2 \frac{90^\circ}{2}}=2r$ so feurbach tangency point gets mapped to mixtilinear tangency point $\implies CF$ is $C$ mixtilinear cevian. Hence $CF,CE$ are isogonals and hence $CE=CF$.