$O_1D \cap OB=P,O_1E \cap OC=Q,OO_1 \cap MN=S$，$S,F,P$ and $S,G,Q$ both are collinear $PQ \parallel BC \parallel FG$，$PQ$ pass through $T$ which is the midpoint of $MN$，$R$ the center of $(POO_1Q)$ $RT \perp PQ$，so $T$ also is the midpoint of $PQ$，the follow is easy......

lym wrote: $S,F,P$ and $S,G,Q$ both are collinear ???

This proof is similar to that given by lym but i will give more details .... Let the lines $ O1 D $ and $ OB $ intersect at $ B' $ and $ O1 E $ and line $ OC $ intersect at $ C' $ and since the line $BB' $ is orthogonal to $ DB'$ and $ BB' $ bisects externally the angle $ CBA $ , the line $ B'C' $ passes midway between $ BC $ and $ DE $ . Let $ K $ be the internal center of homothey of the two circles and since $ N $ and $ M $ are corresponding points in this homothety , S lies on $ NM $ . So $ S $ is the harmonic conjugate of $ A $ for line segment $ O_{1} O $ . Also , lines $ OD $ , $ B' F $ and $ O1B $ are concurrent as cevians in triangle $ OB'O1 $ so the trace of cevian $ B' F $ on $ AO $ is the harmonic conjugate of $ A $ for the line segment $ O_{1} O $ . Similar argument show that the trace of cevian of $ C' G $ on $ A O $ is the harmonic conjugate so the lines $ B'F $ , $ C'G $ , $ NM $ and line $ AO_ {1 } $ are concurrent . Next , that is ye easy then to see that the centre of the circle through $ O _ { 1} $ , $ B' $ , $ O $ and $ C' $ lies on $ O _ {1} O $ since the triangle constructed by drawing line $ B'A' $ and $ C' A' $ which are parallel to $ AB $ and $ AC $ respectively has $ O _{1} $ and $ O $ as their incenter and excenter respectively . So, it is easy to see that the midpoint of $NM $ and the midpoint of $ B'C' $ are the same since $ B' M C' N $ form a parallelogram and so the line joining $ K $ and the midpoint of $ B' C' $ is the median in triangle $ KB' C' $ and finally , $ FG $ is parallel to $ BC $ by the converse of three homothety theorem applied to the line segments $ DE $ , $ BC $ and $ F G $ and this completes the proof then ....

I didn't notice the intersection of $MN$ and $OO_{1}$ makes a harmonic division, lym's solution is nice ! My Solution $Lemma$ In the figure of this problem, assume S: the midpoint of BC, then AS,BD,CE are concurrent. This is easy to prove by Menelause's theorem. It's obvious that FG//BC//DE, also could be proved by Menelause's theorem, so assume T: midpoint of DE, here I wanna use my own notations , $O=I_{a}$ and $O_{1}=I_{p}$ , while $r_{a}$,$r_{p}$ are corresponding radius of these 2 circles. then the question is equal to prove that $MN,I_{a}T,I_{p}S$ are concurrent, since $I_{a}T, I_{p}S$ pass the midpoint of FG by $Lemma$ Apply Desargues Theorem, the question is equal to prove that 3 points $P=I_{a}I_{p} \cap ST,K=I_{p}M \cap SN,U=I_{a}M \cap TN$ , are conlinear. Assume $U_{1}=TN \cap PK,U_{2}=I_{a}M \cap PK$, and $\frac{DE}{BC}= \lambda $, by Menelause Theorem: $\frac{PU_{1}}{U_{1}K} \cdot \frac{KN}{NS} \cdot \frac{ST}{TP}=1$ $\frac{KU_{2}}{U_{2}P} \cdot \frac{PI_{a}}{I_{a}I_{p}} \cdot \frac{ST}{TP}=1$ Assume $a=BC,b=AC,c=AB,c>b$ Then $KN=\frac{c-b}{2a} [(a+b+c)-\lambda (b+c-a)], NS=\frac{c-b}{2}$ and $\frac{PI_{a}}{I_{a}I_{p}}= \frac{a+b+c}{(a+b+c)-\lambda (b+c-a)}$, some trigonometry is needed here but not hard. Then $\frac{PU_{1}}{U_{1}K} \cdot \frac{KN}{NS} \cdot \frac{ST}{TP} \cdot \frac{KU_{2}}{U_{2}P} \cdot \frac{PI_{a}}{I_{a}I_{p}} \cdot \frac{ST}{TP}=\frac{PU_{1}}{U_{1}K} \cdot \frac{KU_{2}}{U_{2}P}=1$ could be proved by simple calculation, and it's done.

We'll use vectors. Let $A$ be the origin, $a,b,c$ be the sidelengths of $\triangle{ABC}$, and $t=DE/BC<1$. Then \[O=\frac{bB+cC}{b+c-a},\quad O_1=\frac{tbB+tcC}{a+b+c}.\]Since $F=BO_1\cap DO$, there exist $r,s$ such that $F=rB+(1-r)O_1=sD+(1-s)O$, i.e. \[F=rB+\frac{t(1-r)b}{a+b+c}B+\frac{t(1-r)c}{a+b+c}C=tsB+\frac{b(1-s)}{b+c-a}B+\frac{c(1-s)}{b+c-a}C.\]Because $A,B,C$ are not collinear, the $B$ and $C$ terms both must be zero, so \[F=\frac{t^2(b+c-a)-t(a+b+c)}{t^2(b+c-a)-(a+b+c)}B+\frac{t(t-1)(bB+cC)}{t^2(b+c-a)-(a+b+c)}.\]By symmetry, we find that \[\frac{F+G}{2}=\frac{t^2(b+c-a)-t(a+b+c)}{t^2(b+c-a)-(a+b+c)}\frac{B+C}{2}+\frac{t(t-1)(bB+cC)}{t^2(b+c-a)-(a+b+c)}.\]Because \[M=\frac{(a-b+c)B+(a+b-c)C}{2a}=\frac{B+C}{2}-\frac{(b-c)(B-C)}{2a}\]and \[N=\frac{t(a+b-c)B+t(a-b+c)C}{2a}=\frac{t(B+C)}{2}+\frac{t(b-c)(B-C)}{2a},\]we can easily verify that $M,N,(F+G)/2$ are collinear, as desired (e.g. show that $(N-M)\times(F+G-2N)=0$).

Another solution : Let CO intersect EO_1 at point X Not hard to prove that angle EXC = 90 Reflect point O_1 wrt OC and OB and get points Y and Q Easy to see that VX || to BC , so QY || BC Let OB and OC intersect DE at points T and Z CO_1 intersect OY at point P BO_1 intersect OQ at point U OE intersect BC at point H YE is anglebissector of CEZ , so CX = XZ and YCO_1Z is rombus PC || YZ , so easy to see that triangles PHC and YEZ are homotetive with center O PH || YE || EO_1 , so triangles PHL and O_1EN are homotetive with center at G PL || NO_1 , so PL is perpendicular to BC Like the same US is perpendicular to BC OQ = OO_1 = OY Easy to see that figures PUBC and YQTZ are homotetiv with center at point O , so UP || QY and OU = OP PUSL is rectangle Use simmetry , so M is midpoint of SL O_1G/O_1P = O_1F/O_1U = NG/NL = NF/NS , so FG || SL SL || FG , so MN goes thru midpoint of FG . done

My solution: Let $ X, Y, Z $ be the midpoint of $ DE, BC, MN $ . Let $ P=OM \cap DE, Q=O_1N \cap BC, V=AN \cap BC, W=MO \cap (O) $ . From homothety we get $ V $ is the tangent point of the incircle of $ \triangle ABC $ with $ BC $ , so we get $ Y $ is the midpoint of $ VM $ and $ A, N, V, W $ are collinear , hence $ O, Y, Z $ are collinear (Similarly, we can get $ O_1, X, Z $ are collinear) . Since $ A, O, O_1 $ all lie on the bisector of $ \angle BAC $ , so from Desargue theorem (for $ \triangle HDB $ and $ \triangle GEC $ ) we get $ GH \parallel BC\parallel DE $ . $ (1) $ Since $ NQMP $ is a rectangle $ \Longrightarrow $ $ P, Z, Q $ are collinear , so from Desargue theorem(for $ \triangle MYO $ and $ \triangle NO_1X $) we get $ O_1Y, OX, MN $ are concurrent . $ (2) $ From $ (1) $ and $ (2) $ we get $ MN $ pass through the midpoint of $ GH $ . Q.E.D