Assuming that "(GIE) and (HDI) are symmetric wrt I" means that (GIE) and (HDI) have the same size, and are tangent at I, this problem is wrong?

wording has been corrected, enjoy / start solving

bumping it is a pity a China TST geo problem to remain unsolved

Let $N$ be the antipode of $A$ wrt. $\omega:=\odot(ABC)$. By angle chasing, $\measuredangle BLC = \measuredangle BLK+\measuredangle KLC = \measuredangle BDK+\measuredangle KEC =\measuredangle(AB,DE)+\measuredangle(DE,AC) = \measuredangle BAC$. So $L \in \omega$. Since $\measuredangle MNL=\measuredangle ANL = \measuredangle ABL = \measuredangle DBL = \measuredangle DKL = \measuredangle MKL$. Hence, $L,K,M,N$ are concyclic. Let $M',N'$ be the reflection of $M,N$ across $L$. We get $D,E,L,M',N'$ are concyclic on $\omega_1$. Let $O,O_1$ be the center of $\omega$ and $\omega_1$. We get $\measuredangle M'O_1L = 2 \measuredangle M'N'L = 2 \measuredangle MNL = \measuredangle AOL$. Hence, $\triangle M'O_1L \sim \triangle LOA$. Let $O_1'$ be a point on $\overrightarrow{MA}$ such that $MO_1'\cdot MA = ML \cdot MM' = ME \cdot MF$. We get $M',L,O_1',A$ are concyclic. So, $\measuredangle LAO = \measuredangle LAO_1' = \measuredangle LM'O_1' = \measuredangle LM'O_1$. So, $O_1',O_1,M'$ are collinear. Moreover, Since $D,E,A,O_1'$ are concyclic and $O_1'A$ is external bisector of $\angle DAE$. So, $O_1',O_1$ lie on the perpendicular bisector of $EF$. Clearly, $M'$ doesn't lie on the perpendicular bisector of $DE$. Therefore, $O_1'=O_1$. We can conclude that $O_1$ lies on $AM$. Lemma: In an isosceles triangle $\triangle ABC$ with $AB=AC$. $D$ lies on $BC$ such that $C$ is midpoint of $BD$. $AD$ intersects $\odot(ABC)$ again at $E$. Then, perpendicular bisector of $CE$ intersects $AD$ again at the reflection of $A$ across $E$. Proof: Let perpendicular bisector of $CE$ intersects $AD$ at $F$. Since $\measuredangle CEF = \measuredangle CBA$. $\triangle CEF \sim \triangle CBA$. Since $-1=(B,D;C,\infty) \overset{A}{=} (B,E;C,A)$. We get $\frac{AE}{EC}=\frac{AB}{BC} = \frac{EF}{EC}$ impiles $AE=EF$. Apply the lemma to $\triangle O_1M'L$. We get $A$ is the midpoint of $O_1O$. Let $F'\ne A$ be a point on $AC$ such $O_1F' = O_1A$. We get $\triangle O_1F'A \equiv \triangle OAB$ and they are isosceles. So $O_1OBF'$ become isosceles trapizoid. $BF' \parallel O_1O \perp BC$. Since $\triangle O_1F'A \sim \triangle O_1ED$ and $\triangle O_1EA \sim \triangle O_1ME$. We get $\triangle O_1EA \cup F' \sim \triangle O_1ME \cap D$. Therefore, $\frac{EF'}{F'A} = \frac{MD}{DE}$ but $\frac{EF}{FM} = \frac{MD}{DE}$. So, $F'F \parallel AM \parallel F'F$. $B,F,F'$ are collinear and perpendicular to $BC$. [asy][asy] import graph; import geometry; size(350); pair A,B,C,D,E,M,N,bruh,bruhh,O,O1,MM,NN,F,FF; A=dir(90); B=dir(200); C=dir(-20); N=dir(-90); O=(0,0); M = B/2+C/2; E = (3+2*sqrt(3))/(5+3*sqrt(3))*A + (2+sqrt(3))/(5+3*sqrt(3))*B; D=extension(A,C,M,E); bruh = D/2+E/2; bruhh = rotate(90, bruh)*E; O1 = extension(bruh,bruhh,A,M); path CC = circle(O1,length(O1-D)); path CCC = circumcircle(A,B,C); pair[] LL = intersectionpoints(CC,CCC); pair L=LL[0]; MM=2*L-M; NN=2*L-N; pair bruhhh =dir(160); FF= extension(bruhhh,B,A,C); F=extension(E,M,bruhhh,B); fill(A--B--C--cycle,0.1*cyan+0.9*white); fill(O1--E--D--cycle,0.8*lightgreen+0.2*white); fill(O1--A--FF--cycle,palegreen); fill(O--A--B--cycle,palegreen); fill(O1--MM--L--cycle,0.8*pink+0.2*white); fill(A--O--L--cycle,0.8*pink+0.2*white); draw(unitcircle); draw(circumcircle(L,D,E),orange); draw(A--B--C--cycle); draw(A--D); draw(M--D); draw(FF--B); draw(N--O1); draw(M--MM); draw(N--NN); draw(MM--O1--L,linewidth(0.2)+red); draw(A--L--O,linewidth(0.2)+red); draw(B--O); draw(FF--O1); draw(D--O1); draw(O1--E); dot("$A$",A,dir(60)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot(E,dir(90)); dot("$E$",D,dir(120)); dot("$M$",M,dir(-45)); dot("$O_1$",O1,dir(0)); dot("$L$",L,dir(-100)); dot("$M'$",MM,dir(190)); dot("$N'$",NN,dir(175)); dot("$F'$",FF,dir(40)); dot("$N$",N,dir(N)); dot("$F$",F,dir(230)); dot("$O$",O,dir(0)); [/asy][/asy]

Simple angle chasing yields that points $A,B,L,C$ are concyclic.Let $T$ be the midpoint of arc $BC$.Now we establish our main claim: Suppose that $\odot(LKM)$ and $\odot(LDE)$ are tangent at $L$,we prove that $K\in \odot(T,TB)$ . The given tangency yields $\angle DLM+\angle ELK=180^{\circ}$,therefore $\angle DLM=\angle ABK$,similarly we have $\angle ELM=180^{\circ}-\angle ACK$. We have $$\begin {aligned}\frac{DM}{EM}&=\frac{[DML]}{[EML]}\\&=\frac{DL\sin DLM}{EL\sin ELM}\\&=\frac{\sin LED}{\sin LDE}\frac{\sin ACK}{\sin ABK}\\&=\frac{\sin LCK}{\sin LBK}\frac{\sin ACK}{\sin ABK}\\&=\frac{BK\sin KLC}{CK\sin KLB}\frac{BK\sin CAK}{CK \sin BAK}\\&=\frac{BK^2}{CK^2}\frac{\sin DEA}{\sin EDA}\frac {\sin CAK}{\sin BAK}\\&=\frac{AD}{AE}\frac{BK^2}{CK^2}\frac{\sin CAK}{\sin BAK}\end{aligned}$$ Since $AM$ is the external bisector of $\angle EAD$,$\frac{DM}{EM}=\frac{AD}{AE}$,therefore $\frac{BK^2}{CK^2}=\frac{\sin BAK}{\sin CAK}$,which establishes the claim.(Trivial computations left as an exercise to the reader) The rest is easy.By angle chasing,$\angle DLE=\angle KLE-\angle KLD=180^{\circ}-\angle{ACK}-\angle {ABK}=180^{\circ}-B $. $\angle LTM=\angle LTA=\angle LBA=\angle LBD=\angle LKD$,indicating that $L,K,M,T$ concyclic. Therefore the radius of $\odot (LED),\odot(LKM)$ is equal can be represented as $\frac{ED}{\cos B}=\frac{KT}{\cos KMB}$. Whlie $KT\cos B=TB\cos B=BM,ED=MF$,we have $MF \cos FMB=BM$,and the result follows.

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