$D,E$ are the midpoints of $BC,CA.$ Then $I \in \odot(ODE)$ $\Longleftrightarrow$ $\angle CIO =90^{\circ}$ $\Longleftrightarrow IO^2+IC^2=OC^2 \Longleftrightarrow R^2-2Rr+IC^2=R^2$ $\Longleftrightarrow \frac{ab(s-c)}{s}=2Rr \Longleftrightarrow 2ab(s-c)= abc \Longleftrightarrow a+b=2c$

Remark: If, additionally, $F$ is the foot of the internal angle bisector of $\angle ACB$ on $AB$, see that $AF=AE$ and $BF=BD$, hence $IE=ID$ and, easily, $CEID$ is cyclic, but $CEOD$ is cyclic as well... Observation: $I$ is the midpoint of $CP$, where $P$ is the second intersection of the internal angle bisector of $\angle ACB$ with the circle $\odot O$. Best regards, sunken rock

Luis González wrote: $I \in \odot(ODE)$ $\Longleftrightarrow$ $\angle CIO =90^{\circ}$ And what if $O\in\lbrace D,E\rbrace$? Then you don't know if circles circumscribed on triangles $CDE,IDE$ are the same (unless you prove it, but you didn't). But certainly $I \in \odot(ODE)$ $\Longleftarrow$ $\angle CIO =90^{\circ}$ which proves the task.

We use barycentric coordinates, with $\triangle ABC$ as the reference triangle. Let $M (1:0:1)$ be the midpoint of $AC$ and $N(0:1:1)$ be the midpoint of $BC$. We have $I = (a:b:c)$ and $$O = (a^2(b^2+c^2-a^2):b(c^2+a^2-b^2):c^2(a^2+b^2-c^2))$$We wish to show that $O$ lies on $(MNI)$. By plugging in $M, N, I$ into the equation for a circle and including the given condition, we get \begin{align*} 2(u+v)=b^2 \\ 2(v+w)=a^2 \\ abc=ua+vb+wc \\ 2c=a+b, \\ \end{align*}from which we get $$(MNI): -a^2yz-b^2zx-c^2zx+\left(\frac{b^2x+a^2y}{2}\right)(x+y+z)=0.$$The rest is just computation to check that $O$ lies on the circle. $\blacksquare$

If $\triangle ABC$ is isosceleces then condition $2AB=AC+BC$ implies that $\triangle ABC$ is equilateral which means that the incentre of $\triangle ABC$, the circumcentre of $\triangle ABC$ coincide and the problem statement is obviously true. Otherwise, WLOG assume that $AC<AB<BC$. Let $M,N$ be midpoints of $AC, BC$ respectively. Obviously $AMON$ is cyclic so it suffices to show $AMIN$ is cyclic. $\triangle AMN$ is a scalene triangle and $I$ belongs to the angle bisector of $\angle MAN$ so it suffices to show $IM=IN$. Let incircle touch $AC,BC$ at points $D,E$ respectively. We can easily see that $IM=IN \Longleftrightarrow MD=NE$. Also, $MD=NE \Longleftrightarrow (s-c) - \frac{b}{2} = \frac{a}{2} - (s-c)$ which is true because of the condition $2AB=AC+BC$.

Switch to $A$-indexing. Clearly $M,N$, the midpoints of $AB,AC,$ in addition to the circumcenter $O$ and $A$ are concyclic with diameter $OA.$ Thus, it suffices to show that $I$ lies on $(AMN)$. The general equation of a circle is $$-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)=0$$for some constants $u,v,w$. To find the constants $u,v,w$ which determine $\Omega$, we can plug in each of $A,M,N$ in the general form and solve for $u,v,w$. Note that the equation of a circle is homogenous so we can use homogenized coordinates here. Plugging in $A=(1,0,0)$ gives $u=0$. Plugging in $M=(1:0:1)$ gives $u+w=\frac{b^2}2$. Plugging in $N=(1:1:0)$ gives $u+v=\frac{c^2}2$. Plugging in these constants in the general equation, we find that $\Omega$ has equation $$-a^2yz-b^2zx-c^2xy+(x+y+z)\left(\frac{c^2}2\cdot y+\frac{b^2}2\cdot z\right)=0.$$Hence, using Barycentric Power of a Point, we can compute $$\text{Pow}_{\Omega}(I)=-(a+b+c)(abc)+(a+b+c)\left(\frac{c^2}2\cdot b+\frac{b^2}2\cdot c\right)$$since $I=(a:b:c)$. But this is $0$ since $a=\frac{b+c}2$. Done.