We wish to prove that $M$ lies on the radical axis of the two circles. With $AC = b$, $AB = c$ and $O_B$ being the circumcenter of $E_BI_CI$, if we find an expression for the power $MO_B^2 - R_{E_BI_CI}^2$ which is symmetric with respect to $b$ and $c$, then the analogue for the other circle shall be the same and we will be done. Observe that $\angle I_CO_BE_B = 2\angle I_CIE_B = 90^{\circ}$ and hence $O_BI_CCE_B$ is cyclic. As $I_CE_B = R\sqrt{2}$, $I_CO_B = R, E_BO_C = R$, we deduce by Ptolemy for $I_CO_BE_BC$ (with $\angle I_CO_BE_B = \angle I_CCE_B = 90^{\circ}$) that $O_BC = \frac{CI_C + CE_B}{\sqrt{2}}$. To reduce $MO_B$ to the computation of nicer segments, note that by the Cosine Law in triangle $MCO_B$ we have (with $a=BC$ and $\angle ACB = \gamma$) $O_BM^2 = O_B^2C + \frac{a^2}{4} - a \cdot O_BC \cdot \cos\left(45^{\circ} + \frac{\gamma}{2}\right)$. Hence (using the Pythagorean theorem on $E_BCI_C$) $$ O_BM^2 - R_{E_BI_CI}^2 = \frac{(CI_C+CE_B)^2}{2} + \frac{a^2}{4} - a \cdot \frac{CI_C + CE_B}{\sqrt{2}} \cdot \cos\left(45^{\circ} + \frac{\gamma}{2}\right) - \frac{CI_C^2 + CE_B^2}{2} $$$$ = CI_C \cdot CE_B + \frac{a^2}{4} - \frac{a}{\sqrt{2}}(CI_C + CE_B) \cdot \cos\left(45^{\circ} + \frac{\gamma}{2}\right). $$Now from $\triangle DAC \sim \triangle ABC$ we have $\frac{CI_C}{CI} = \frac{DA}{AB} = \cos\gamma$ and from $\triangle IE_BC$ with $\angle ICE_B = 90^{\circ}, \angle E_BIC = 45^{\circ}$ we get $CE_B = CI$ and so the above expression becomes $$ CI^2\cos\gamma + \frac{a^2}{4} - \frac{a}{\sqrt{2}}\cdot CI \cdot (1+\cos\gamma)\cdot \cos\left(45^{\circ} + \frac{\gamma}{2}\right). $$ We now compute $CI = \sqrt{\frac{ab(a+b-c)}{a+b+c}}$ (from the angle bisector formula), $\cos \gamma = \frac{b}{a}$, $\sin\frac{\gamma}{2} = \sqrt{\frac{1 - \cos \gamma}{2}} = \sqrt{\frac{a-b}{2a}}$, as well as $\cos\left(45^{\circ} + \frac{\gamma}{2}\right) = \sqrt{\frac{1+\cos(90^{\circ}+\gamma)}{2}} = \sqrt{\frac{1-\sin \gamma}{2}} = \sqrt{\frac{a-c}{2a}}$, so having in mind $a = \sqrt{b^2+c^2}$ the main expression becomes $$ \frac{b^2(a+b-c)}{a+b+c} + \frac{a^2}{4} - \frac{a+b}{2} \sqrt{\frac{b(a+b-c)(a-c)}{a+b+c}}. $$To argue that the latter is symmetric with respect to $b$ and $c$, we may remove the summand $\frac{a^2}{4}$ and multiply with $a+b+c$ and so work instead with $b^2(a+b-c) - \frac{a+b}{2}\sqrt{b(a-c)(a+b-c)(a+b+c)}$.

It suffices to show $M$ lies on the radical axis of the two circles. To do, for a point $X$ define: $$ f(X)=\mathrm{Pow}_{\odot E_{C}II_{B}}{\left(X\right)}-\mathrm{Pow}_{\odot ABC}{\left(X\right)} $$which is a linear function. We then have $f(C)=CI \cdot CE_{C}=ab$ and also, using $\triangle ADB \cap I_{B} \sim \triangle CAB \cap I$ we get: $$ f(B)=BI_{B} \cdot BI=BI^2 \cdot \frac{c}{a}=\frac{c}{a} \cdot \frac{BI}{BE_{B}} \cdot \underbrace{BI \cdot BE_B}_{ac}=\frac{c^2 (s-b)}{s} $$Then we have, by linearity: \begin{align*} f(M)&=\frac{f(B)+f(C)}{2} \\ &=\frac{abs+c^2(s-b)}{2s} \\ &=\frac{ab(a+b+c)+c^2(a+c-b)}{4s} \\ &=\frac{a^2b+a\left(b^2+c^2\right)+abc+c^3-bc^2}{4s} \\ &=\frac{a^3+\cancel{bc^2}+b^3+c^3+abc-\cancel{bc^2}}{4s} \tag{$a^2=b^2+c^2$} \\ &=\frac{a^3+b^3+c^3+abc}{4s} \end{align*}We thus have: $$ \mathrm{Pow}_{\odot E_{C}II_{B}}{\left(M\right)}=f(M)+\mathrm{Pow}_{\odot ABC}{\left(M\right)}=\frac{a^3+b^3+c^3+abc}{4s}-\frac{a^2}{4} $$which is symmetric under $b \leftrightarrow c$ so $\mathrm{Pow}_{\odot E_{C}II_{B}}{\left(M\right)}=\mathrm{Pow}_{\odot E_{B}II_{C}}{\left(M\right)}$ which is what we wanted.

Let $O_1$ and $O_2$ be centers of $II_CE_B$ and $II_BE_C$ and Let $R_1$ and $R_2$ be their radius. Claim $ : O_1I_CCE_B$ and $O_2I_BBE_C$ are cyclic. Proof $:$ Note that $\angle I_CO_1E_B = 2\angle I_CIE_B = \angle 90$. we prove the other one with same approach. Note that we need to prove $MO_1^2 - R_1^2 = MO_2^2 - R_2^2$. $MO_1^2 = O_1C^2 + \frac{a^2}{4} - a . O_1C . \cos{(45 + \frac{C}{2})}$ $R_1^2 = \frac{I_CE_B}{2\sin{45}}^2 = \frac{I_CE_B^2}{2}$ $O_1C^2 = CI_C.CI + R_1^2$ so $MO_1^2 - R_1^2 = CI_C.CI + \frac{a^2}{4} - a . O_1C . \cos{(45 + \frac{C}{2})}$ so we need to prove $CI^2 - pow_{BI_BI_CC}(I) - a . O_1C . \cos{(45 + \frac{C}{2})} = BI^2 - pow_{BI_BI_CC}(I) - a . O_2B . \cos{(45 + \frac{B}{2})}$ or $CI^2 - a.\frac{CI_C+CE_B}{\sqrt{2}}.\cos{(45 + \frac{B}{2})} = BI^2 - a.\frac{BI_B+BE_C}{\sqrt{2}}.\cos{(45 + \frac{C}{2})}$ and Note that $\frac{CI_C}{CI} = \frac{b}{a}$ and $\frac{BI_B}{BI} = \frac{c}{a}$ and $\frac{CE_B}{BE_C} = \frac{BI}{CI}$ then just expand both sides and we're Done. Any synthetic by the way ?

any synthetic?

Here's a synthetic proof: Lukaluce wrote: Let $ABC$ be a right-angled triangle with $\angle BAC = 90^{\circ}$. Let the height from $A$ cut its side $BC$ at $D$. Let $I, I_B, I_C$ be the incenters of triangles $ABC, ABD, ACD$ respectively. Let also $E_B, E_C$ be the excenters of $ABC$ with respect to vertices $B$ and $C$ respectively. If $K$ is the point of intersection of the circumcircles of $E_CII_B$ and $E_BII_C$, show that $KI$ passes through the midpoint $M$ of side $BC$. Claim: $I_BI_C \parallel E_BE_C$. Proof: Consider the spiral sim $\sigma$ at $D$ sending $C \to A$ and $A \to B$. Then $\sigma$ sends $I_C$ to $I_B$. Hence, $$ \triangle DCI_C \sim \triangle DAI_B \implies \triangle DCA \sim \triangle DI_CI_B $$As $\angle I_CDC = 45^\circ$, hence $\angle (I_BI_C,AC) = 45^\circ$. But its direct that $\angle (E_BE_C,AC) = 45^\circ$. Hence $I_BI_C \parallel E_BE_C$. $\square$ Now note points $E_B,E_C,B,C$ are concyclic (they lie of circle with diameter $E_BE_C$). So Reim's Theorem gives that points $I_B,I_C,C,B$ are concyclic. Now ignore all points in the figure and only look at points $$ I,B,C,I_B,I_C,E_B,E_C $$Invert at $I$. Our problem just becomes equivalent to the following well-known Ceva's Theorem application: Let $ABC$ be a triangle and $D,E,F$ be points on sides $BC,CA,AB$, respectively. Suppose $AD,BE,CF$ concur. If $D$ is the midpoint of segment $BC$, then $EF \parallel BC$.

guptaamitu1 wrote: Our problem just becomes equivalent to the following well-known Ceva's Theorem application Could you please clarify this? How exactly does inverting make the two problems equivalent? Thank you in advance!

After inversion intersection of two circles becomes the intersection of two lines. Further, after inversion line joining points $I_BI_C$ and $E_BE_C$ both become parallel to $BC$. We obtain the required configuration. Also, the official solution does something similar but without inversion.