China Girls' Mathematical Olympiad

Let $P$ be the intersection of line $AM$ with $EF$. Join $AT$, $BM$, $BP$, $BT$, $CM$, $CT$, $ET$, $TP$. As shown in Fig. 7.2.

As $BF$ is tangent to $\odot O_2$ at $F$, we have $\angle BFT=\angle FET$.

As $\odot O_1$ is tangent to $\odot O_2$ at $T$, we have $\angle MBT=\angle FET$.

Hence, $\angle MBT=\angle BFM$. So $△MBT$ and $△MFB$ are similar, therefore $M{B}^2=MT\cdot MF$. The same argument gives $M{C}^2=MT\cdot MF$.

Now again from that $\odot O_1$ is tangent to $\odot O_2$ at $T$, we have $\angle MAT=\angle FET$. So $A$, $E$, $P$ and $T$ are concyclic, which implies that $\angle APT=\angle AET$. As $AE$ is tangent to $\odot O_2$ at $E$, we have $\angle AET=\angle EFT$. Thus, $\angle MPT=\angle PFM$, and $△MPT$ is similar to $△MF$. Therefore, $M{P}^2=MT\cdot MF$.

From the above argument, we have $MC=MB=MP$, which means that $M$ is the excenter of $△BCP$. So $\angle FBP=\frac{1}{2}\angle CMP$. Meanwhile, $\angle CMP=\angle CDA=\angle ABF$, and we have $\angle FBN=\frac{1}{2}\angle ABF$. Hence, $\angle FBN=\angle FBP$, i.e., $P$ and $N$ coincide. $\square$

Fig. 7.2