My solution : Let $ T $ be the point on the segment $ AM $ such that $ MB=MT=MC $ . From $ MT=MB \Longrightarrow \measuredangle ATB=90^{\circ}+\tfrac{1}{2} \measuredangle AMB=\measuredangle AI_bB \Longrightarrow T \in \odot (ABI_b) $ . Similarly, we can prove $ T $ lie on $ \odot (ACI_c) \Longrightarrow T \equiv \odot (ABI_b) \cap \odot (ACI_c) \in AM $ . Q.E.D

Let $X$ be the second intersection of the circumcircle of $ABI_b$ and $ACI_c$. Note that $\angle XBI=\angle IAX=\angle IAB=\angle BXI$ $\implies$ $BI=IX.$ Also, $\angle BIM=90^{\circ}+\frac{BAM}{2}$ and $\angle ABX=\angle ABI-\angle XBI=\frac{\angle B}{2}-\frac{\angle BAM}{2}$. $\angle XIM $ $=\angle AIM-\angle XIA $ $=(90^{\circ}+\frac{\angle B}{2})-\angle XBA $ $=(90^{\circ}+\frac{\angle B}{2})-(\frac{\angle B}{2}-\frac{\angle BAM}{2}) $ $=90^{\circ}+\frac{\angle BAM}{2} $ $=\angle BIM$ Hence, by SAS, $\triangle BIM \cong \triangle XIM$ so $\angle IXM=\angle IBM=\angle IBA$. Because $\angle AXI=180^{\circ}-\angle ABI$, $\angle AXI+\angle IXM =180^{\circ},$ as desired. $\blacksquare$

zschess wrote: Let $ ABC $ be a triangle, and let $M$ be midpoint of $BC$. Let $ I_b $ and $ I_c $ be incenters of $ AMB $ and $ AMC $. Prove that the second intersection of circumcircles of $ ABI_b $ and $ ACI_c $ distinct from $A$ lies on line $AM$. This is the so--called "sparrow's lemma": if the circle passes through th points $A$ and $I$ (icenter) in intersects lines $AB$ and $AC$ in points $X$ and $Y$, then $BX+CY=BC$. Using this lemma, consider the intersiction points $T_1$ and $T_2$ of two circumcircles with $AM$. Hence $$0 + MT_1=MB=MB=MT_2+0.$$So $T_1=T_2$ and circumcircles pass through one point $T=T_1=T_2$ on $AM$.

Let $ ABI_b $ meet AM at S. $\angle MSB = \angle 180 - \angle ASB = \angle 180 - \angle AI_bB = \angle 90 - \angle AMB/2$ so $\angle SBM = \angle SMB = \angle 90 - \angle AMB/2$ so $MB = MS = MC$. $\angle AI_cC = \angle 180 - (\angle 90 - \angle AMC/2) = \angle 180 - \angle CSM = \angle ASC$ so $ ACI_cC $ is cyclic.

Inversion works here. Invert respect circle centered at $A$.