GeoMetrix wrote:

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -19.43752933627648, xmax = 32.95131572780579, ymin = -22.91737803351104, ymax = 11.67967094711988; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((1.88,-6.82)--(15.2,-6.7), linewidth(0.4) + rvwvcq); draw((15.2,-6.7)--(1.948187283488726,4.409082587313549), linewidth(0.4) + rvwvcq); draw(circle((5.47643671228076,-3.212986968017347), 3.574467648278019), linewidth(0.4) + dtsfsf); draw((1.88,-6.82)--(5.47643671228076,-3.212986968017347), linewidth(0.4) + wvvxds); draw(circle((6.747870334297736,-9.23935115462224), 8.825349861185424), linewidth(0.4) + wvvxds); draw(circle((8.521543675938103,-4.711348029129539), 6.968250536085029), linewidth(0.4) + sexdts); draw((2.2692084696477464,-1.63485327018204)--(1.90203496433293,-3.191281833747048), linewidth(0.4) + wrwrwr); draw((1.948187283488726,4.409082587313549)--(11.659998503978759,-16.571322036789724), linewidth(0.4) + wvvxds); draw((15.2,-6.7)--(11.659998503978759,-16.571322036789724), linewidth(0.4) + wrwrwr); draw((2.2692084696477464,-1.63485327018204)--(7.772785268839476,-0.47371657695024716), linewidth(0.4) + wvvxds); draw((7.772785268839476,-0.47371657695024716)--(5.5086378167960826,-6.787309569218054), linewidth(0.4) + wvvxds); draw((7.772785268839476,-0.47371657695024716)--(4.336270300841782,-0.7498879530434238), linewidth(0.4) + wrwrwr); draw((-1.7486378167960828,-6.852690430781947)--(11.659998503978759,-16.571322036789724), linewidth(0.4)); draw((2.2692084696477464,-1.63485327018204)--(-1.7486378167960828,-6.852690430781947), linewidth(0.4)); draw((-1.7486378167960828,-6.852690430781947)--(7.772785268839476,-0.47371657695024716), linewidth(0.4)); draw((7.772785268839476,-0.47371657695024716)--(11.659998503978759,-16.571322036789724), linewidth(0.4)); draw((1.948187283488726,4.409082587313549)--(1.88,-6.82), linewidth(0.4) + rvwvcq); draw((2.2692084696477464,-1.63485327018204)--(11.659998503978759,-16.571322036789724), linewidth(0.4) + wvvxds); draw((-1.7486378167960828,-6.852690430781947)--(1.88,-6.82), linewidth(0.4) + wrwrwr); draw((2.2692084696477464,-1.63485327018204)--(4.336270300841782,-0.7498879530434238), linewidth(0.4) + wrwrwr); /* dots and labels */ dot((1.948187283488726,4.409082587313549),dotstyle); label("$A$", (2.0963271088950033,4.739849912657041), NE * labelscalefactor); dot((1.88,-6.82),dotstyle); label("$B$", (2.0282896477728185,-6.486331172503435), NE * labelscalefactor); dot((15.2,-6.7),dotstyle); label("$C$", (15.329613297159941,-6.350256250259066), NE * labelscalefactor); dot((5.47643671228076,-3.212986968017347),linewidth(4pt) + dotstyle); label("$I$", (5.6002563566875185,-2.9483831941498306), NE * labelscalefactor); dot((5.5086378167960826,-6.787309569218054),linewidth(4pt) + dotstyle); label("$D$", (5.634275087248612,-6.5203499030645276), NE * labelscalefactor); dot((7.772785268839476,-0.47371657695024716),linewidth(4pt) + dotstyle); label("$E$", (7.913530034841801,-0.19286601870135017), NE * labelscalefactor); dot((1.90203496433293,-3.191281833747048),linewidth(4pt) + dotstyle); label("$F$", (2.0282896477728185,-2.914364463588738), NE * labelscalefactor); dot((-1.7486378167960828,-6.852690430781947),linewidth(4pt) + dotstyle); label("$Q$", (-1.6117145222640668,-6.588387364186712), NE * labelscalefactor); dot((2.2692084696477464,-1.63485327018204),linewidth(4pt) + dotstyle); label("$K$", (2.4024956839448346,-1.34950285777849), NE * labelscalefactor); dot((11.659998503978759,-16.571322036789724),linewidth(4pt) + dotstyle); label("$I_A$", (11.791665318806333,-16.283725574098032), NE * labelscalefactor); dot((4.336270300841782,-0.7498879530434238),linewidth(4pt) + dotstyle); label("$L$", (4.47763824817147,-0.465015863190089), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
IMO the title doesnt suit the prob.
Proof: Let $I_A$ be the $A$ -excentre. Its well known(ISL 2002 G7) that $K=I_AD \cap \omega$ . Now we'll use a phantom approach . Let $\ell$ be the line through $F$ parallel to $BI$ . and let $\ell \cap BC=Q'$ We just need to show that $Q' \in \odot(KEG)$ . Firstly define $\ell \cap AI=L$ . Notice that $\angle LQ'C=\angle IBC=\angle IKC=\angle LKC \implies LQ'I_AC$ cyclic. Now notice that $\Delta FAL \cong \Delta EAL \implies \angle AFL=\angle AEL$ . But notice that $\angle AEL=\angle AFL=\angle ABI=\angle LQ'C\implies LEQ'C$ cyclic. This combined with the fact that $LQ'I_AC$ is cyclic $\implies LQ'CI_AE$ is cyclic. So now we just need to show that $K\in \odot(LQ'CI_AE)$ But notice that $\angle I_AEQ'=\angle I_ACQ'=90-\frac{\angle C}{2}$ and $\angle Q'I_AE=\angle Q'CE=\angle C \implies I_AQ=I_AE$ . FInally $\angle EKD =\angle CDE=\angle I_ACD=\angle I_AEQ'=\angle I_AQ'E\implies K \in \odot(I_AQ'E)$ as desired. $\blacksquare$

Well there was no need of phantom point approach
Here's how you can finish it easily:
$\triangle AEI_A\equiv \triangle AFI_A\Rightarrow I_AE=I_AF\Rightarrow I_A$ is circumcenter of $QFE$ .Since $\angle QI_AE=C$ we get $\angle QFE=180-\frac {C}{2}$ and we are done.