Let $I$ be a incenter of $\triangle ABC$. Denote by $2\alpha, 2\beta, 2\gamma$ angles $\angle A, \angle B, \angle C$ of the $\triangle ABC$. Let $\odot (ABI)$ intersect side $\overline{BC}$ in $E'$. Claim. $E' \equiv E$ Proof. We have that $\angle KE'C = 180^\circ - \angle AE'B = 180^\circ - \angle AIB = 90^\circ - \gamma \Longrightarrow \angle E'AC = 90^\circ - \gamma \Longrightarrow CE'=CA$. $\square$ Analogously, for point $D$. [asy][asy] size(10cm); import olympiad; pair A,B,C,K,F,E,D,T, G, P; A = dir(110); B = dir(170); C = dir(10); K = incenter(A,B,C); E = intersectionpoints(B--C, circle(C, 1.53))[0]; D = intersectionpoints(B--C, circle(B, 1))[0]; G = intersectionpoint(B--K, A--E); F = intersectionpoint(C--K, A--D); P = 11*K-10*A; T = intersectionpoints(A--P, circumcircle(K,D,E))[1]; draw(A--B--C--A); draw(A--D); draw(A--E); draw(circumcircle(K,D,E), dashed); draw(circumcircle(A,B,K), dotted); draw(circumcircle(A,C,K), dotted); draw(A--T); draw(F--E); draw(F--T); draw(G--D); draw(G--T); draw(T--E); draw(T--D); draw(F--G); draw(B--K); draw(C--K); draw(K--E); draw(K--D); label(scale(0.8)*"$A$", A, dir(110), filltype = Fill(white)); label(scale(0.8)*"$B$", B, dir(200), filltype = Fill(white)); label(scale(0.8)*"$C$", C, dir(-20), filltype = Fill(white)); label(scale(0.8)*"$K$", K, dir(60), filltype = Fill(white)); label(scale(0.8)*"$T$", T, dir(-70), filltype = Fill(white)); label(scale(0.8)*"$D$", D, dir(-40), filltype = Fill(white)); label(scale(0.8)*"$E$", E, dir(200), filltype = Fill(white)); label(scale(0.8)*"$G$", G, dir(160), filltype = Fill(white)); label(scale(0.8)*"$F$", F, dir(50), filltype = Fill(white)); dot(A); dot(B); dot(C); dot(E); dot(D); dot(K); dot(G); dot(F); dot(T); [/asy][/asy] Observe that $\angle AGK = \angle BGA + \angle BAG = \beta + (2\alpha + \gamma - 90^\circ) = \alpha = \angle KAC = \angle KDC$, so $G$ lie on the circumcircle of the $\triangle KDE$. By the same argument we have that $F,G \in \odot (KDE)$. Claim. $I \equiv K$ Proof. Just notice that $\angle AIE = 180^\circ - 2\beta$ and $\angle ADE = \angle ACE + \angle CAE = 2\gamma + (2\alpha - (90^\circ - \beta)) = 90^\circ - \beta$, so $\angle AIE = 2\angle ADE$. Similarly, $\angle AID =2\angle AED$ and the conclusion is follows. $\square$ Set $T = \odot (KDE) \cap \overline{AK}$. To finish the proof we just notice that $\angle GTD = 180^\circ - \angle GKD = 180^\circ - \angle GKA = \angle GKT = \angle GDT$, so $GT=GT$ and similarly $FT=FE$. $\blacksquare$

Slave wrote: Let $I$ be a incenter of $\triangle ABC$. Denote by $2\alpha, 2\beta, 2\gamma$ angles $\angle A, \angle B, \angle C$ of the $\triangle ABC$. Let $\odot (ABI)$ intersect side $\overline{BC}$ in $E'$. Claim. $E' \equiv E$ Proof. We have that $\angle KE'C = 180^\circ - \angle AE'B = 180^\circ - \angle AIB = 90^\circ - \gamma \Longrightarrow E'AC = 90^\circ - \gamma \Longrightarrow CE'=CA$. $\square$ Analogously, for point $D$. Claim. $I \equiv K$ Proof. Just notice that $\angle AIE = 180^\circ - 2\beta$ and $\angle ADE = \angle ACE + \angle CAE = 2\gamma + (2\alpha - (90^\circ - \beta)) = 90^\circ - \beta$, so $\angle AIE = 2\angle ADE$. Similarly, $\angle AID =2\angle AED$ and the conclusion is follows. $\square$ Set $T = \odot (KDE) \cap \overline{KL}$. To finish the proof we just notice that $\angle GTD = 180^\circ - \angle GKD = 180^\circ - \angle GKA = \angle GKT = \angle GDT$, so $GT=GT$ and similarly $FT=FE$. $\blacksquare$ fantastic!!! amazing dude

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.

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.

A very nice and enjoyable problem! Here's the solution. We first deal with $K$. Claim : $K$ is the incenter of $\triangle ABC$ Proof : Note that, \[\measuredangle EKA = 2\measuredangle EDA = 2\measuredangle BDA = \measuredangle DBA = \measuredangle EBA\]which implies that $ABEK$ is cyclic. Similarly, we obtain that $AKDC$ is also cyclic. Now, \[\measuredangle ABK = \measuredangle AEK = \measuredangle KAE = \measuredangle KBE\]as a result of $ABEK$ being cyclic. Similarly, we can also show that $\measuredangle ACK = \measuredangle KCD$ which proves that $K$ is indeed the incenter of $\triangle ABC$ as claimed. Now, we can also make the following observations. Claim : Points $G$ and $F$ lie on the circle $(KDE)$. Proof : First, note that \[\measuredangle GBE = \measuredangle KBE = \measuredangle ABK\]and \[\measuredangle BEG = \measuredangle BEA = \measuredangle BKA\]which implies that $\triangle BEG \sim \triangle BKA$. This then gives us that $\measuredangle EGB = \measuredangle KAB$ from which we conclude that \[\measuredangle EGB = \measuredangle KAB = \measuredangle CAK = \measuredangle EDK\]which implies that $EDKG$ is cyclic. Similarly, we can also show that $EDFK$ is cyclic, which proves the claim. Claim : $GA=GD$ and $FA=FE$. Proof : We have that \begin{align*} 2\measuredangle DAE &= 2 \measuredangle DAB + 2 \measuredangle CAE + 2\measuredangle BAC\\ &= \measuredangle DBA + \measuredangle ACE + 2\measuredangle BAC\\ &= (\measuredangle CBA + \measuredangle BAC + \measuredangle ACB) + \measuredangle BAC\\ &= \measuredangle BAC\\ &= 2\measuredangle BKC \end{align*}where the last equality is since $K$ is the incenter of $\triangle ABC$. Then, it follows that $\measuredangle DAE = \measuredangle BKC$. Thus, \[\measuredangle DAG = \measuredangle DAE = \measuredangle BKC = \measuredangle GKF = \measuredangle GDF\]from which it is quite clear that $GA=GD$. Similarly, we can show that $FA=FE$ as well. Now comes the key claim of the problem. Claim : Let $X= \odot (G,D) \cap (KED)$ and $Y= \odot (G,D) \cap (KED)$. Then, $A-K-X$ and $A-K-Y$. Proof: We prove one - the other is entirely similar. First, observe that \[\measuredangle AGK = \measuredangle EGB = \measuredangle EDK = \measuredangle KED = \measuredangle KGD\]Now, \[2\measuredangle AXD = \measuredangle AGD = 2\measuredangle KGD = 2\measuredangle KXD\]from which it follows that $\measuredangle AXD = \measuredangle KXD$, which implies that $A-X-D$ as desired. Thus, the intersections of the circles $c_1$ and $c_2$ and $c_1$ and $c_3$ both lie on $\overline{AK}$ which implies that $X=Y$ and thus, the circumcircle of triangle $KDE$ , the circle with center the point $F$ and radius $FE$ and the circle with center $G$ and radius $GD$ concur on a point which lies on the line $AK$.