It is easy using complex coordinates. Set $\omega_1$ as the unit circle. We have $c=\frac{a^2}{b}$ because $AB=AC$. I will not write out the whole computation in detail, because it is very routine if you are familiar with complex coordinates. We can calculate $d=2b-\frac{a^2}{b}$, $e=\frac{a(a+2b)}{2a+b}$, $f=\frac{b^3}{a^2}$, $g=\frac{a^2+ab-b^2}{a}$. Then we can calcuate the intersecting point of $CE$ and $BG$.(Actually, $BG$ is tangent to $\omega_1$). The intersecting point is $\frac{2ab}{a+b}$. This point is where the tangent lines of $\omega_1$ at $A$ and $B$ intersect. Therefore it is on the circumcircle of $AOB$. There is probably a nice synthetic solution, although I haven't given it much thought. Edit: The synthetic solution is as simple as I thought it would be, here is a sketch: Simple angle chasing shows that $BG$ is tangent to $\omega_1$. Let $K$, $M$ be the midpoints of segments $AB$ and $AC$, then $AD//BM$, so $\angle DAB=\angle ABM=\angle ACK$. Now it is easy to see that $CE$ is the symmedian. Therefore the intersecting point of $CE$ and $BG$ is where the tangent lines of $\omega_1$ at $A$ and $B$ intersect.