IMO 1999 Bucharest geometry problem

Two circles $G_1$ and $G_2$ are contained inside the circle G, and are tangent to G at the
distinct points M and N, respectively. $G_1$ passes through the center of $G_2$ . The line
passing through the two points of intersection of $G_1$ and $G_2$ meets G at A and B.
The lines MA and MB meet $G_1$ at C and D, respectively.
Prove that CD is tangent to $G_2$

Here it is a proof without (many) words.
Homotopy of center M which maps C in A transform $G_1$ in $G$ .so $D$ passes in $B$ so line $CD$ passes in line $AB$ and $AB$ is parallel to $CD$ . As $AB$ is orthogonal to $O_1O_2$ we have the same relation $CD$ orthogonal to $O_1O_2$ .
Inversion of pole B which maps P in Q , transforms $G_1$ in $G_1$ , transforms $G_2$ in $G_2$ and transforms $G$ to common tangent line $ED$ so $$O_1O_2ED$ is a trapezoid with $O_1O_2=O_1D$ so $O_2F=HD=O_2E=r_1$ so $CD$ is tangent to $G_2$