Let $O$ be the cicumcenter of $\triangle ABC$ $OM=\frac{AH}{2}$ Let $L$ be the point of tangency of $\omega_2$ and circumcircle of $\triangle ABC$ Hence, $O-M-L$ are collinear $\implies ML = OL-OM = R-\frac{AH}{2}$ where $R$ denotes circumradius of $\triangle ABC$ Let $X$ be the center of $\omega_1 \implies X$ is midpoint of $AH$ Therefore, $AX=OM , AX \parallel OM$ (as both are perpendicular to $BC$) Hence, $AXMO$ is a parallelogram $\implies XM=AO=R$ Hence, distance between centers of $\omega_1$ and $\omega_2$ is $R$ Radius of $\omega_1 = \frac{AH}{2}$, Radius of $\omega_2 = R-\frac{AH}{2}$ Hence, Radius of $\omega_1 +$ Radius of $\omega_2 = R =$ Distance between centers of $\omega_1$ and $\omega_2$ Hence, the circles $\omega_1$ and $\omega_2$ are tangent to each other.