If $AB = AC$, then the graph is symmetric about the bisector of $\angle BAC$ and the conclusion is obvious. So we may assume that $AB > AC$. As shown in Fig. 8.2, let $L$ be the midpoint of $BC$, line $O_1L$ intersecting with $FG$ at $R$, and $O_1N$ be extended to intersect with $BC$ at $K$. Draw line $AT$ that is perpendicular to $BC$ at $T$ and intersects with line $DE$ at $S$. Connecting $AO$, obviously $O_1$ is on the segment $AO$. By Menelaus' theorem, we have

Fig. 8.2

$$\frac{O_1 F}{F B} \cdot \frac{B D}{D A} \cdot \frac{A O}{O O_1} = 1, \quad \frac{O_1 G}{G C} \cdot \frac{C E}{E A} \cdot \frac{A O}{O O_1} = 1. \qquad \textcircled{1}$$

Since $DE \parallel BC$, we have $\frac{BD}{DA} = \frac{CE}{EA}$. So $\frac{O_1 F}{F B} = \frac{O_1 G}{G C}$, which means that $FG \parallel BC$. Then $\frac{FR}{GR} = \frac{BL}{CL} = 1$. Therefore, $R$ is the midpoint of $FG$. We now only need to prove that $M, R, N$ are collinear. For this purpose, by the inverse of Menelaus' theorem, we only need to prove that

$$\frac{O_1 R}{R L} \cdot \frac{L M}{M K} \cdot \frac{K N}{N O_1} = 1. \qquad \textcircled{2}$$

Since $FR \parallel BL$, we have $\frac{O_1 R}{R L} = \frac{O_1 F}{F B} = \frac{O O_1}{A O} \cdot \frac{A D}{D B}$ (the second equality is justified by ①). So we only need to prove that

$$\frac{O O_1}{A O} \cdot \frac{A D}{D B} \cdot \frac{L M}{M K} \cdot \frac{K N}{N O_1} = 1. \qquad \textcircled{3}$$

Since $O_1 K \perp DE$, $O M \perp BC$, $A T \perp BC$, $DE \parallel BC$, then lines $O_1 K$, $O M$, $A T$ are parallel. By the theorem of dividing the segments into proportional by parallel lines, we have $\frac{O O_1}{A O} = \frac{M K}{M T}$. Substituting it into ③, we then only need to prove

$$\frac{A D}{D B} \cdot \frac{L M}{M T} \cdot \frac{K N}{N O_1} = 1. \qquad \textcircled{4}$$

Since $DE \parallel BC$, $K N \perp DE$, $S T \perp BC$, quadrilateral $K N S T$ is a rectangle and then $K N = S T$. Furthermore, from $D S \parallel B T$ we have $\frac{A D}{D B} = \frac{A S}{S T}$. Substituting these results into ④, we now only need to prove

$$\frac{LM}{MT} = \frac{NO_1}{AS}. \qquad \textcircled{5}$$

Let $BC = a$, $AC = b$, $AB = c$. We have

$$BM = \frac{a+b-c}{2} \text{ (the property of an inscribed circle)}, \quad BL = \frac{a}{2},$$
$$BT = c \cos \angle ABC = c \cdot \frac{a^2 + c^2 - b^2}{2ac} = \frac{a^2 + c^2 - b^2}{2a}.$$
Then
$$\frac{LM}{MT} = \frac{BL - BM}{BT - BM} = \frac{\frac{c-b}{2}}{\frac{c^2 - b^2 + a(c-b)}{2a}} = \frac{a}{a+b+c}.$$
On the other hand,
$$\frac{NO_1}{AS} = \frac{\frac{2S_{\triangle ADE}}{AD + DE + AE}}{\frac{2S_{\triangle ADE}}{DE}} = \frac{DE}{AD + DE + AE} = \frac{a}{a+b+c}.$$
Therefore, ⑤ holds. The proof is complete.

Let the radii of $\odot O$ and $\odot O_1$ be $r$ and $r_1$, respectively. Obviously, $O_1$, $O$ and $A$ are collinear.
$$\left. \begin{array}{l} DE // BC \Rightarrow \frac{AB}{BD} = \frac{AC}{CE} \\[2ex] \frac{OF}{FD} = \frac{S_{\triangle BOO_1}}{S_{\triangle DBO_1}} = \frac{\frac{1}{2}\left(AB \sin \frac{A}{2}\right) \cdot OO_1}{\frac{1}{2}r_1 \cdot BD} \\[2ex] \frac{OG}{GE} = \frac{S_{\triangle OO_1C}}{S_{\triangle EOO_1}} = \frac{\frac{1}{2}\left(AC \sin \frac{A}{2}\right) \cdot OO_1}{\frac{1}{2}r_1 \cdot CE} \\[2ex] \Rightarrow FG \parallel DE \parallel BC. \end{array} \right\} \Rightarrow \frac{OF}{FD} = \frac{OG}{GE}$$

Connect $ON$ and extend it to intersect with $BC$ at $K$. If $\angle ABC = \angle ACB$, then by symmetry, the proposition holds. So we may assume that $\angle ABC < \angle ACB$ in the following.
We connect $OM, O_1M, OB, MD, DO_1$, respectively (see Fig. 8.3). Since $O_1N \parallel OM$, we have

Fig. 8.3

$$S_{\triangle ONM} = S_{\triangle MOO_1} = \frac{1}{2} r \cdot OO_1 \cdot \sin \frac{C-B}{2}, \qquad ①$$
$$\frac{OG}{GE} = \frac{OF}{DF} = \frac{S_{\triangle BOO_1}}{S_{\triangle BDO_1}} = \frac{\frac{1}{2}BO \cdot OO_1 \cdot \sin \frac{C}{2}}{\frac{1}{2} \cdot BD \cdot DO_1 \cdot \sin \frac{B}{2}}$$
$$= \frac{r \cdot OO_1 \cdot \sin \frac{C}{2}}{r_1 \cdot BD \cdot \cos \frac{B}{2}} \qquad ②$$
$$(r = BO \cdot \cos \frac{B}{2}, r_1 = DO_1 \cdot \sin \frac{B}{2}),$$
$$S_{\triangle DMN} - S_{\triangle MEN} = \frac{1}{2}NK \cdot DN - NE \\ = \frac{1}{2}BD \cdot \sin B \cdot \left( r_1 \cot \frac{B}{2} - r_1 \cot \frac{C}{2} \right). \quad ③$$
From ② and ③, we have
$$\begin{align*} & \frac{OG}{GE} S_{\triangle DMN} - S_{\triangle MEN} \\ &= \frac{1}{2} r_1 \cdot BD \cdot \sin B \cdot \left( \cot \frac{B}{2} - \cot \frac{C}{2} \right) \cdot \frac{r \cdot OO_1 \cdot \sin \frac{C}{2}}{r_1 \cdot BD \cdot \cos \frac{B}{2}} \\ &= r \cdot OO_1 \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} \cdot \left( \cot \frac{B}{2} - \cot \frac{C}{2} \right) \\ &= r \cdot OO_1 \cdot \sin \frac{C-B}{2}. \end{align*}$$
Combining it with ①, we get
$$\frac{OG}{GE} (S_{\triangle DMN} - S_{\triangle MEN}) = 2S_{\triangle MON}.$$
Since $\frac{OG}{GE} : 2 = \frac{OG}{OE} : \left(\frac{DF}{OD} + \frac{EG}{OE}\right)$, then
$$\frac{OF}{OD} \cdot S_{\triangle MND} - \frac{OG}{OE} \cdot S_{\triangle MEN} = \left(\frac{DF}{OD} + \frac{EG}{OE}\right) \cdot S_{\triangle MON}, \\ \frac{OF \cdot S_{\triangle MND} - DF \cdot S_{\triangle MON}}{OD} = \frac{OG \cdot S_{\triangle MEN} + EG \cdot S_{\triangle MON}}{OE}. \quad ④$$
On the other hand,
$$S_{\triangle NMG} = \frac{S_{\triangle MEN} \cdot OG + EG \cdot S_{\triangle MON}}{OE}.$$
$$\text{Therefore, } S_{\triangle NMG} = \frac{OF \cdot S_{\triangle MND} - DF \cdot S_{\triangle MON}}{OD}.$$
In the same way,
$$S_{\triangle NMF} = \frac{OF \cdot S_{\triangle MND} - DF \cdot S_{\triangle MON}}{OD}.$$
By ④, we have $S_{\triangle NMG} = S_{\triangle NMF}$. Therefore, $MN$ divides segment $FG$ equally.
The proof is complete.