, 2003

We consider two cases:

* First case: The circles $(O_1), (O_2)$ touch each other externally at $M$.
Let $xy$ be the common tangent at $M$ of $(O_1), (O_2)$. Since $CA$ and $My$ are tangents to $(O_1)$ at $C$ and $M$, we have $\angle FCA = \angle CMy$. But $\angle CMy = \angle FMx$, hence $\angle FCA = \angle FMx$. As $\angle FMx = \angle FAM$ it follows that $\angle FCA = \angle FAM$.
The triangles $MFA$ and $AFC$ have: $\angle MFA = \angle AFC$ and $\angle FAM = \angle FCA$, so they are similar. Therefore $\frac{MF}{FA} = \frac{AF}{FC}$ t.c. $FM \cdot FC = FA^2$. But $FM \cdot FC = P_M(O_1)$ (the power of $M$ with respect to the circle $(O_1)$) $= FO_1^2 - R_1^2$, where $R_1$ is the radius of $(O_1)$, it follows that $FO_1^2 - FA^2 = R_1^2$. Analogously, we have $EO_1^2 - EA^2 = R_1^2$. So
$$FO_1^2 - FA^2 = EO_1^2 - EA^2 = R_1^2.$$
As $D$ lies on the line $EF$, it implies that $DO_1^2 - DA^2 = R_1^2$, i.e.

$$DO_1^2 - R_1^2 = DA^2 \quad (1)$$
Since $DA$ is the tangent to $(O_2)$ at $A$, $DA^2 = PD/(O_2) = DO_2^2 - R_2^2$ (2), where $R_2$ is the radius of $(O_2)$.
From (1) and (2), we get $DO_1^2 - R_1^2 = DO_2^2 - R_2^2$, or $PD/(O_1) = PD/(O_2)$. It shows that $D$ lies on the radical line of $(O_1)$ and $(O_2)$.

* 2^d case: The circles $(O_1), (O_2)$ touch each other internally at $M$.
The proof in this case is analogous to the proof in the 1st case.