As shown in figure , it is known that $ABCD$ is parallelogram, $A,B,C$ lie on circle $\odot O_1$, $AD$ and $BD$ intersect $\odot O$ at points $E$ and $F$ respectively, $C,D,F$ lie on circle $\odot O_2$, $AD$ intersects $\odot O_2$ at point $G$. If the radii of circles $\odot O_1$, $\odot O_2$ are $R_1, R_2$ respectively, prove that $\frac{EG}{AD}=\frac{R_2^2}{R_1^2}$.