Let $A$-excircle be the unit circle and line $BI_a$ be the $x$-axis. Let $E,F,R$ are the tangency points of $A-$excircle and $\overline{AC},\overline{AB},\overline{BC}$ small letters are complex coordinate of big letters! $a=\frac{2ef}{e+f}$, since $D$ is reflect of $A$ over $\overline{BI_a}$ so $d=\bar{a}=\frac{2}{e+f}$ $K=\overline{I_aD}\cap \overline{EF}$ so $k=\frac{ef}{e^2f^2+1}$ by concurrency of lines formula. Let $O$ be circumcenter of $\triangle EKD$ so $o=$ $\begin{bmatrix*} d & d\bar{d} & 1 \\ e & e\bar{e} & 1 \\ k & k\bar{k} & 1 \\ \end{bmatrix*}$ $\div$ $\begin{bmatrix*} d & \bar{d} & 1 \\ e & \bar{e} & 1 \\ k & \bar{k} & 1 \end{bmatrix*}$ $=\frac{e(2ef^2+e+f)}{(e+f)(e^2f^2+1)}$ Now notice $b=\frac{2}{\bar{f}+\bar{r}}=\frac{2}{\bar{f}+f}$ and $c=\frac{2}{f+\bar{e}}$ Let $M$ be midpoint of $BC$. So $m=\frac{b+c}{2}=\frac{2f^2e+e+f}{f^3e+f^2+fe+1}$ Now what's left is proving $\frac{o}{m}\in \mathbb{R}$ Notice by what we got above $\frac{o}{m}=\frac{e(f^3e+f^2+fe+1)}{(e+f)(e^2f^2+1)}=\frac{e(ef+1)(f^2+1)}{(e+f)(e^2f^2+1)}$ But by $\bar{f}=\frac{1}{f}$ and $\bar{e}=\frac{1}{e}$ now it's obvious that $\frac{o}{m}=\overline{\frac{o}{m}}$

Gaussian_cyber wrote: I have a solution by complex numbers. Any synthetic without inversion? Can you post it, please?

Let the second tangent from $M$ touch the excircle at $S$. Let the incircle touch $BC$ at $R$. Lemma 1: $A,R,S$ are collinear.

Lemma 2: $AR$ and $I_AM$ are parallel. This follows lemma 1 by noting that $M$ is the midpoint of $RG$ and $I_A$ is the midpoint of $GG'$. Lemma 3: $DI_ACA$ is cyclic. $$\angle{C}=2\angle{AI_AB}=\angle{AI_AD}=\angle{DCA}$$So the third lemma holds as well. Let $T$ be the reflection of $E$ about $MI_A$. We will now prove that triangles $ASI_A$ and $DTI_A$ are congruent. $$\angle{TI_AS}=\angle{GI_AE}=\angle{C}=\angle{DI_AA}$$by the Lemma 3. So $\angle{TI_AD}=\angle{SI_AA}$ and we have $I_AT=I_AS\ ,\ I_AD=I_AA$ So that is proven. Now by the congruency above, we have $\angle{TDI_A}=\angle{SAI_A}=\angle{LI_AM}=\angle{TEK}$ so $EKTD$ is cyclic and we are done since $ET$ is the common chord of the two circles and thus $I_AM$ which is it's perpendicular bisector, also includes the center of the other circle!

Another finish using the lemmas above. (and the figure) Let the second intersection of $w_a$ and DKE be T.Let the circumcentere of $(DTKE)$ be O. we need to prove that $O,M,I_a$ are collinear.since $OI_a$ is perpendicular to $ET$ we need to prove $I_aM$ is perpendicular to $ET$ since $AS$ is parallel to $I_aM$ we need to prove $AS$ is perpendicular to ET and cause $AI_a$ is perpendicular to EF we need to prove $\angle{SAI_a}=\angle{LET}$ The same congruent and the fact that $(DTKE)$ is cyclic would tell us : $\angle{LET}=\angle{TDK}=\angle{TDI_a}=\angle{SAI_a}$ .so we are done.

Am I the only one find this problem super hard. Here is a probably an unexpected way to prove it. Let $G = I_AE \cap BC$, $O_1$ be the circumcenter of $\triangle DKE$. The somehow magic claim is that the homothety of center $I_A$ sending $\odot (I_ABC)$ to $\odot I_ADG$ sends $M$ to $O_1$ ! Claim 1: $D,E,K,G$ are concyclic. This followed by $\angle I_AEK = \frac{1}{2} \angle A = \angle I_ADG $. Claim 2: $\odot I_ABC$ is tangent to $\odot I_ADG$ at $I_A$. Let $\ell$ be the tangent line of $\odot I_ABC$ at $I_A$, which is perpendicular to $AI_A$. Hence $\angle (\ell, I_AG) = \angle (I_AA,AE) = \frac{1}{2} \angle A = \angle I_ADG$. Now consider the homothety $\mathcal{H}$ of center $I_A$ sending $\odot I_ABC$ to $\odot I_ADG$. We use $X'$ to denote the image of $X$ under $\mathcal{H}$, i.e. $M'$ is the image of $M$, etc. Claim 3: $M'D = M'G$ Let $O'$ be the circumcenter of $\triangle I_ADG$, then $O'M' \perp B'C' // BC=DG$, therefore $O'M'$ is indeed the perpendicular bisector of $DG$. Claim 4: $M'G = M'E$ We show that $B'G \perp AC$. Indeed $\angle GB'C' = \angle GI_AC' = \angle GI_AC = 90^\circ - \angle BCI_A = 90^\circ - \angle B'C'I_A$, hence $B'G \perp AC$. Let $T = B'G \cap AC$, basic angle chasing shows that $M'T \perp GE$, indeed $\angle (M'T, GE) = \angle (SI_A, GB') = 90^\circ$. Finally we show that $TG =TE$. Note that $\angle CTG = 90^\circ = \angle GEC$, we have $G,T,C,E$ are concyclic. Hence $\angle TEG = \angle TCG = \angle I_ACB = \angle ECI_A = \angle EGT$. Hence $TG= TE$ implying that $M'G=M'E$. This finishes the proof as $M'$ is indeed the circumcenter of $D,E,G,K$.

Sometimes, instead solving the incircle version really helps. Replace $D$ with $A'$ and redefine it as the $A$-excircle touch point to $BC$. Let $M$ be the midpoint of $BC$, $P$ be the foot of perpendicular from $M$ to $EF$, $Q=DI_{A}\cap EF$ and $R=AI_{A}\cap BC$. Claim 1: $QR\parallel AD$. Proof: Let $S=EF\cap BC$. It is well-known that $SI_{A}\perp AD$ (usually the incircle version appears.). But in $\triangle SQR$, $I_{A}$ is the orthocenter, so $SI_{A}\perp QR$ and the claim follows. Claim 2: Let $EI_{A}$ meet $BC$ at $T$. Then $T$ lies on $(KA'E)$. Proof : This is easy angle chasing. Just notice that $$\measuredangle TEK=\measuredangle I_{A}EF=\measuredangle I_{A}AF=-\measuredangle I_{A}A'D=\measuredangle TA'K.$$ From this, we see that $\triangle I_{A}EK$ and $\triangle I_{A}A'T$ are inversely similar. Let $N$ be the midpoint of $EF$. Then the pairs $(N,D)$ and $(Q,R)$ are correspondent in the two triangles. $P$ and $M$ are also correspondent because $$\frac{DR}{RM}=\frac{AQ}{QM}=\frac{NQ}{QP},$$where the ratios are written in directed lengths. (Here we used the fact that $A,Q,M$ are collinear. This is also known but usually the incircle version.) This means that if the perpendicular bisector of $EK$ and $TA'$ meet $I_{A}M$ at $O_{1}$ and $O_{2}$, $\frac{I_{A}O_{1}}{O_{1}M}=\frac{I_{A}O_{2}}{O_{2}M}$, which would imply that $O_{1}=O_{2}$. In particular, as $E,K,T,A'$ are concyclic, this is their circumcenter, so we're done.

Solution using cross ratio. Let $T = I_A \cap BC$, $S = EF \cap BC$, $R = FT \cap DE$. We have $\measuredangle KET = \measuredangle FEM = \measuredangle FAI_A = \measuredangle I_ADB = \measuredangle KDT$, hence $E, F, D, T$ are concyclic. By Brocard Theorem, we are enough to prove that $SR \perp I_AM$. By the property of quadrilateral, we have $S (D, F; R, I_A) = -1$. And $SD \perp AH$, $SF \perp AI$, $SI_A \perp ANa$, $AGe // MI_A$, hence we only need to show that $A(H, I; Na, Ge) = -1$, which is well-known.

It was tempting to bash.

let $D'= I_aE \cap BC$ $T,T'$ be the in-touch , ex-touch point with $BC$ and let $Y$ the second tangent from $M$ to $(I_a)$ claim: $D'EKD$ is cyclic $\angle D'EF=\angle D'EK=180-\angle I_ADD'=180-\angle KDD'$ $\blacksquare$ now let $E'$ be a point on $(D'EKD)$ such that $EE' || T'Y$ we'll prove $E' \in (I_a)$ with few angle chasing we have (it's well-known that $AT || I_aM$) $$\angle FEE'=\angle E'DK=\angle YAI_a=\angle AI_aM$$so in order to prove that $E' \in (I_a)$ we have to prove $\angle T'YE+\angle YAI_a=FEY$ $\angle FEY=\angle YFA=180-\angle FYA-\angle YAF=\angle T'YE+\angle YAI_a$ and we win

Let $O$ be center of $DKE$ and $M$ be midpoint of $BC$ and $I$ be incenter and $P$ be touch point of incenter and $BC$ and $AP$ meet $\Omega$ at $S$. Note that $O$ and $I_a$ are center of $DKE$ and $\Omega$ so we need to prove $MI_a$ is perpendicular to $DKE$ and $\omega$. Let perpendicular at $E$ to $MI_a$ meet $\Omega$ at $E'$ so we need to prove $DE'KE$ is cyclic. Claim $: DI_aCA$ is cyclic. Proof $:$ Note that $\angle ADC = \angle IBC = \angle II_aC = \angle AI_aC$. Claim $: MS = MZ$. Proof $:$ Note that $AP$ passes through antipode of $\Omega$ so $\angle ZSP = \angle ZSA = \angle 90$ and it's well known that $Z$ is midpoint of $P,Z$ so $MZ = MS$. Claim $: E'DI_a$ and $SAI_a$ are congruent. Proof $:$ Note that $DI_a = AI_a$ and $E'I_a = SI_a$ and $\angle E'I_aS = \angle ZI_aE = \angle C = \angle AI_aD \implies \angle DI_aE' = \angle AI_aS$. $\angle E'DK = \angle E'DI_a = \angle SAI_a = \angle MI_aA$. Note that $\angle MI_a \perp E'E$ and $AI_a \perp EF$ so $\angle MI_aA = \angle E'EF = \angle E'EK$ so $\angle E'DK = \angle E'EK$ so $DE'KE$ is cyclic as wanted.

Let $M$ be the midpoint of $\overline{BC}$ and $O$ as the center of $(DKE)$. Denote the incircle of $\triangle ABC$, the intouch point, and the extouch point on $\overline{BC}$ as $I$, $T$, and $T_A$, respectively. Finally, let $\overline{AT}$ meet the excircle at points $X$ and $Y$, where $X$ lies in between $A$ and $Y$. We will rewrite the end condition as follows: Points $O$ and $I$ are the centers of $(DKE)$ and the excircle, respectively. If we denote $E'$ as the point on the excircle such that $\overline{MI_A} \perp \overline{EE'}$, and $E'$ also lies on $(DKE)$, we will be done by radical axis properties. Claim 1: Points $A$, $C$, $D$, and $I_A$ are concyclic. Proof: Notice that \[\angle ADC = \angle IBC = \angle II_AC = \angle AI_AC. \ \square\] Claim 2: $MX=MT_A$. Proof: It is well-known that $Y$ is the antipode of $T_A$ with respect to the excircle, so $\angle T_AXY = \angle T_AXT = 90^\circ$. It is also well-known that the midpoint of $\overline{TT_A}$ is $M$, so $MT=MT_A=MX$. $\square$ Claim 3: $\triangle E'DI_A \cong \triangle XAI_A$. Proof: Clearly, $AI_A=DI_A$, and $EI_A=XI_A$. Then, we also have \[\angle E'I_AX = \angle T_AI_AE = \angle ACB = \angle AI_AD,\] where the last step follows from Claim 1. This implies $\angle E'I_AD = \angle AI_AX$, so we are done by SAS congruence. $\square$ Observe that $\overline{AT} \parallel \overline{MI_A}$, so \[\angle E'DK = \angle E'DI_A = \angle AI_AX = \angle XAI_A = \angle MI_AA.\] Note that $\overline{MI_A} \perp \overline{EE'}$ and $\overline{AI_A} \perp \overline{EF}$. Thus, \[\angle E'DK = \angle MI_AA = \angle E'EF = \angle E'EK,\] proving that $E'$ lies on $(EDK)$, as desired. $\square$

Solved with MathLuis who completely carried. Really nice and instructive problem. Let $\omega_A$ be the $A-$excircle and $X$ the tangency point of $\omega_A$ with $BC$. We can start off by locating $D$. Claim : Point $D$ lies on $\overline{BC}$. Further, $DI_ACA$ is cyclic. Proof : We let $D'$ be the point on the extension of $\overline{BC}$ beyond $B$ such that $AB=BD'$. Notice that then in triangles $\triangle I_ABA$ and $\triangle I_ABD$, $AB=BD'$ and $BI_A$ is a common side. Further, \[\measuredangle I_ABD' = \measuredangle I_ABC = \measuredangle FBI_A = \measuredangle I_ABA\]which proves that $\triangle I_ABA \cong \triangle I_ABD'$. Thus, $D'$ must be the reflection of $A$ across $I_AB$ which implies that $D'=D$ and indeed $D$ lies on $\overline{BC}$ as claimed. Now it is easy to see that \[\measuredangle DI_AA =2 \measuredangle BI_AA = 2 (\measuredangle I_ABF + \measuredangle BAI_A) = \measuredangle CBF + \measuredangle BAC = \measuredangle BCA = \measuredangle DCA\]which implies the rest of the claim. Now, we are in a position to attack the problem. Let $O$ be the center of $(DKE)$ and $M$ the midpoint of $BC$. We also let $L$ be the reflection of $E$ across $I_AM$. Let $S$ be the $A-$intouch point and second tangency point (which does not lie on $\overline{BC}$) from $M$ to $\omega_A$ respectively. It is well known that $AR \parallel MI_A$ and $A-R-S$. From the definition it follows that $LI_A=EI_A$ so $L$ lies on $\omega_A$. Further, $DI_A=AI_A$. Also note that the above reflection maps $X$ to $S$. Thus, arc $LS$ maps to arc $XE$ which implies that $\measuredangle LI_AS = \measuredangle EI_AX$. But, since $XCEI_A$ is clearly cyclic, we have that $\measuredangle EI_AX = \measuredangle BCA$. Thus, $\measuredangle LI_AS = \measuredangle BCA$. Now, we simply note that \begin{align*} \measuredangle LI_AD + \measuredangle AI_AS &= \measuredangle FI_AD + \measuredangle LI_AF + \measuredangle AI_AF + \measuredangle FI_AS\\ &= \measuredangle FI_AD + \measuredangle LI_AF + \measuredangle AI_AF + \measuredangle FI_AL + \measuredangle LI_AS\\ &= \measuredangle FI_AD + \measuredangle AI_AF + \measuredangle LI_AS\\ &= \measuredangle 90 + \measuredangle FAI_A + \measuredangle ACB + \measuredangle AI_AF + \measuredangle BCA\\ &= 0 \end{align*}and thus, $\measuredangle LI_AD = \measuredangle AI_AS$ which implies that $\triangle LI_AD \cong \triangle SI_AA$. Now, we note that since $AI_A \perp EF$ and $MI_A \perp ES$, it follows that $\measuredangle AI_AM = \measuredangle FEL$. But, note that this implies \[\measuredangle KEL = \measuredangle FEL = \measuredangle AI_AM = \measuredangle I_AAS = \measuredangle I_ADL = \measuredangle KDL\]which implies that $L$ indeed lies on $(KDE)$. Thus, $LE \perp OI_A$ (radical axis) which implies that $O-M-I_A$ as desired. Remark : There are lots of claims hiding inside this configuration. I noticed at first that $I_AE \cap BC$ lies on $(DKE)$ and $EF \cap BC$ lies on $(DEI_A)$. One can also quite easily show that $DI_A \cap AB$ lies on $(DKL)$ once the problem statement is proved.