Belarusian Mathematical Olympiad

Since $B_1$ is symmetric to $A$ with respect to the midpoint of $BC$, $AB{B}_{1}C$ is a parallelogram, so $B{B}_1\parallel AC$ and $B{B}_1=AC$. Likewise $AC{D}_{1}D$ is a parallelogram, $D{D}_1\parallel AC$ and $D{D}_1=AC$. Therefore the triangle $C{D}_1{B}_1$ is obtained from the triangle $ADB$ by the transfer along the line $AC$ by the length of the segment $AC$. Let $O$ be the center of $\omega$ and $O_1$ be the circumcenter of the triangle $C{D}_1{B}_1$. Then $O{O}_1\parallel AC$. Since the line connecting the centers of two intersecting circles is perpendicular to their common chord, $O{O}_1\perp CG$ which means $AC\perp CG$. Therefore $\angle ACG=90^{\circ }$, i. e. $AG$ is a diameter of $\omega$.



Second solution:

Let $K$, $L$ and $M$ be the midpoints of the sides $BC$, $CD$ and $CA$ respectively. Since the triangle $KLM$ is obtained from the triangle $BDA$ by a homothety with the center $C$ and the coefficient $1/2$, its circumcircle ${\omega }_1$ is obtained from $\omega$ in the same way: $H_C^{0.5}(\omega )={\omega }_1$. The center $C$ of the homothety lies on $\omega$, so it lies on ${\omega }_1$ as well. Since the triangle $B_1{D}_{1}C$ is obtained from the triangle $KLM$ by a homothety with the center $A$ and the coefficient $2$, its circumcircle ${\omega }_2$ is obtained from ${\omega }_1$ in the same way: $H_A^2({\omega }_1)={\omega }_2$. Consider the intersection points of $\omega$ and ${\omega }_2$, namely, how they are obtained under the composition $H_A^2\circ H_C^{0.5}$, translating $\omega$ into ${\omega }_2$. One of them, the point $C$ is obtained from the point $A$, indeed: $H_A^2(H_C^{0.5}(A))=H_A^2(M)=C$. The second point is obtained from the point $E$, lying on $\omega$ diametrically opposite to $C$ since $O=H_C^{0.5}(E)\in {\omega }_1$ and $H_A^2(O)\in {\omega }_2$. Whence $H_A^2(O)=G$, so $AG$ is a diameter.

the inverse transformation: from $\{C,G\}=\omega \cap H_A^2({\omega }_1)$, it follows that $$\{H_A^{0.5}(C),H_A^{0.5}(G)\}=H_A^{0.5}(\omega )\cap {\omega }_1=H_A^{0.5}(\omega )\cap H_C^{0.5}(\omega ).$$ Now, the intersection points of $H_A^{0.5}(\omega )$ and $H_C^{0.5}(\omega )$ are obviously the middle $M$ of the segment $AC$ and the center $O$ of the circle $\omega$. It remains to make the transformation $H_C^2(\omega )$ and get the desired points $A$ and $E$.