Insimilicenter $A$ of $(I_b) \sim (I_c)$ is also center of negative inversion that transforms them into each other. By conformity, $\omega$ is transformed into the common external tangent $\tau_A$ of $(I_b),(I_c)$ different from $BC.$ The center of $\omega$ is on the perpendicular form $A$ to $\tau_A,$ but $\tau_A$ is antiparallel to $BC$ WRT $AB,AC$ $\Longrightarrow$ it is on the A-circumdiameter of $\triangle ABC$ $\Longrightarrow$ $\omega$ is tangent to the circumcircle $(O)$ of $\triangle ABC$ $\Longrightarrow$ $A$ is exsimilicenter of $\omega \sim (O).$ Hence, if $AM,AN$ cut $(O)$ again at $M',N',$ we have $M'N' \parallel MN \equiv BC$ $\Longrightarrow$ arcs $BM',$ $CN'$ are equal $\Longrightarrow$ $\angle BAM=\angle CAN.$

Grade 6? I am speechless.

haha, most probably it is Grade 9

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=524295&hilit=Saint+Peterburg