This question was also from 2022 P1 Azerbaijan BMO TST, proposed by Greece

Boring. Observe that $K$ is the incenter of $ABC$ and let the circumcircle of $KDE$ intersect $AK$ at $T$. Angle chase to show that $TKGE$ is cyclic and hence $TDFKGE$ is cyclic. Then finish the chase by showing that $KF$ is the external angle bisector of $\angle EKT$, implying $FE = FT$ (and analogously $GD = GT$).

Claim: $F,G$ lie on $\omega_1$. Proof: Observe that $BG \bot AD$ as $\overline{BGK}$ is the perpendicular bisector of $AD$. We then have: $$\angle KGA=90^{\circ}-\angle GAF=90^{\circ}-\angle EAD=\angle KDE$$which shows that $G \in \omega_1$ and similarly $F \in \omega_1$. As $G$ lies on the perpendicular bisector of $AD$, $A$ lies on $\omega_3$ and similarly $A$ lies on $\omega_{2}$. Let $X$ be the reflection of $A$ in line $FG$ then $X \in \omega_{2},\omega_{3}$. Also observe $FG$ is antiparallel to $ED$ wrt $\angle EAD$ so $AK \bot FG$ and hence $X \in AK$. We will now show $X \in \omega_1$ which will complete the problem. Using $AD \bot BK$ and $AE \bot CK$: $$\angle GKF=180^{\circ}-\angle GAF=180^{\circ}-\angle FXG$$so, as $G,F \in \omega_1$, we have $X \in \omega_1$ also as needed.

Also Greece National Olympiad 2022 P1 https://artofproblemsolving.com/community/c6h2790100p24533111 Note that $ABD$ is isosceles so $BK$ is angle bisector. with same approach for $CK$ we have $K$ is incenter of $ABC$. $\angle AEB = \angle 90 + \frac{\angle C}{2} = \angle AKB \implies AKEB$ is cyclic. with same approach $AKDC$ is cyclic. $\angle GEK = \angle AEK = \angle EAK = \angle GDK \implies GEDK$ is cyclic. with same approach $FDEK$ is cyclic. Let $KGEDF$ meet $AK$ at $S$. $\angle KSG = \angle KDG = \angle KAG \implies AG = SG \implies S$ lies on circle with center $G$. with same approach $G$ lies on circle with center $F$. we're Done.

Note that $X$ be intersection point of $\omega_1$ and $\omega_2$ , We need to show $KEDX$ is cyclic and $A,K,X$ are collinear Let $\angle AEK = z$ , $\angle KED = y$ and $\angle ADK = x$. Then by angle-chasing (using $KA = KE = KD$ and $BA = BD , CA=CE$) we see that $ABEK$ and $AKDC$ are cyclic, Also $\angle KGE + \angle KDE = 180$ , $\angle KFD + \angle KED = 180$.Hence $F$ and $G$ lies on $\omega_1$.By connecting $G$ and $D$ we see $\angle GAD = z+x = \angle GDA$ , Thus $GA=GD$ and $A\in \omega_3$, Similarly $FA = FE$ and $A \in\omega_2$. $\angle AGK= y = \angle KGD$ hence $GK$ bisects $\angle AGD$ , Let $T$ be intersection of $GK$ and $AD$, Then $T$ is midpoint of $AD$ and $AD\bot GK$, Similarly define $S$ on the segment $AE$. $TS$ is midsegment of $AED$ , so $\angle ATS = x + y$ and $\angle AST = z + y$. $\triangle DFC \implies \angle DFC = y = \angle SFT$ , $STGF$ cyclic $\implies \angle TFG = z+y \implies \angle SFG=z = \angle GFK(1)$ LET US PROVE $A,K,X$ are collinear : $AX$ radical axis of $\omega_2$ and $\omega_3$ $\implies AX\bot FG$. Hence we need to show $AK\bot GF$, let $R$ be intersection of $AK$ and $GF$. $KFED \implies \angle FED = y-x \implies \angle FEK= x$ , $GEKF \implies \angle FGK = x \implies \angle ARG = 180-(x+y+z) = 90$(becouse of sum of angles in $ABC$ ) , Hence $A,K,X$ are collinear , from $\triangle GXA$ $\implies \angle GXA= z = \angle GXK(2)$ From $1$ and $2$ $GKFX$ is cyclic $\implies$ $KEDX$ cyclic , so we are done (I hope I did not do any mistake during writing)

Observe that $CK$ bisects $\angle ACE$ and $BK$ bisects $\angle ABD$ due to $K$'s being on $AE$ and $AD$'s perpendicular bisectors. Trivially $G,F$ are on $\omega_1.$ Let $X$ be $AK\cap \omega_1$ then \[\angle GXD=\angle GED = 90^\circ - \tfrac12 \angle ACB = 180^\circ - (90^\circ + \tfrac12 \angle ACB)=180^\circ-\angle AKB=\angle GKX=\angle GDX\]so $\omega_3$ passes through $X$. Similarly, so does $\omega_2$. We are done.

Here is my solution. lemma 1: K is the incenter of ABC. This is simple .We draw the perpendicular lines to AB and BC from K. Then we prove two triangles are congruent. Thus K is on bisector of B. Similarly we prove K lies on bisector of C. lemma2:DFKGE is cyclic First note that $\angle KDE$=A/2.Since K is the incenter of ABC it can be easily proved that $\angle KGA$=A/2.So KDEG is cyclic. Similarly it can be shown KDEF is cyclic. So our claim is proved. With our lemmas and some angle chasing it is easy to prove that$\angle AFE$=A and $\angle FEA$=$\angle FAE$=90-A/2.Therefore A lies on w2.Similarly A lies on w3. Call the intersection point of w2 and w3 T. We have :FE=FT,AG=GT. So FGA and FGT are congruent triangles.$\angle FTG$=$\angle FAG$ =90-A/2. K Is the incenter of ABC. So$\angle FKG$=90+A/2.Therefore FKGT is cyclic. According to lemma2 we have KDET is cyclic. It remains to prove that A,K and T are collinear. In AKF triangle we'd have $\angle AKF$=90+B/2. $\angle FKT$=$\angle FGT$=$\angle FGA$=$\angle FDE$=90-B/2.SO A,K and T are collinear. QED.

Note that $K, C$ are on the perpendicular bisector of $AE$. So $CK \bot AE$. Let $AK$ intersect $(KED)$ at $X$. Proving $FE=FX$ will be enough(symmetrically $GD=GX$). First notice that by $KEXD$ cyclic $\implies$ $\angle KED=\angle KDE=\angle DXK=\angle KXE$. $\angle CEF=\angle CAF$ so $\angle FEK=\angle FAK=\angle FDK$ $\implies$ $KFDE$ cyclic. So $\angle KAF=\angle KDF=\angle KXF$ $\implies$ $FX=FA=FE$ done.