I prove it by using Menelaus Th. Join $DB,EB,DE,GB$ since $\omega_1 , \omega_2$has diameter on $AC$ ,that is obviously that $\angle GDB=\angle GEB= \frac{\pi}{2}$, so $G,D,B,E$ are concyclic , by easy angle chasing : $\angle DAB=\angle GED,\angle O_2CE=\angle O-2EC=\angle GEF$ , Hence :$\angle DEF=\angle DAB+\angle ECB$ , by the same way , we can also get : $\angle FDE=\angle DAB+\angle ECB=\angle DEF$ , Hence $DF=FE$ , Denote the intersection of $FG$ and $AC$ be $M$ . Then ,to triangle $FMO_2$ and line $GEC$ , by Menelaus Th. :$\frac{CO_2}{CM}*\frac{MG}{GF}*\frac{FE}{EO_2}=1$ (1), to triangle $FMO_1$ and line $GDA$ ,by Menelaus Th. : $\frac{AO_1}{AM}*\frac{MG}{GF}*\frac{FD}{DO_1}=1$ (2), (1)/(2) WE GET :$\frac{AM}{CM}=1$ [Notice that 1.$DF=FE$ ; 2. $CO_2=O_2E,AO_1=O_1D$] hence : $ AM=MC $ . Done !

Dear Mathlinkers, 1. D', E' the second points of intersection of DO1, EO2 with "omega" and H the point of intersection of AE' and CD'. 2. According to Reim's theorem (or a angle chasing) AD // CD' and CE // AE'. 3. The quadrilateral AFCH being a parallelogram, FH goes through the midpoint of AC. 4. According to Pascal's theorem applied to the hexagon AE'ECD'DA, we are done… Sincerely Jean-Louis

Note that $\sphericalangle{EDF}=\sphericalangle{DAC}+\sphericalangle{ACE}=\sphericalangle{FED}$, therefore $FD=FE$. As furthermore $O_1B=O_1D$ and $O_2B=O_2E$, $\odot(BED)$ is the incircle $\tau$ of $\triangle{O_1O_2F}$. As $AG\perp BD$ and $BE\perp CG$, $G\in\odot(BED)$. Moreover, as $\sphericalangle{BDG}=\frac{\pi}{2}$, $G$ is the point opposite of $B$ on $\tau$, implying that $M=FG\cap O_1O_2$ is the touchpoint of an excircle with $O_1O_2$. Therefore, as $B$ is the touchpoint of the incircle with $O_1O_2$, it is well-known that $O_1M=O_2B$ and $O_2M=O_1B$ and hence $AM=AO_1+O_1M=AO_1+O_2C=O_2C+O_2M=MC$, as required.

Martin N. wrote: Note that $\sphericalangle{EDF}=\sphericalangle{DAC}+\sphericalangle{ACE}=\sphericalangle{FED}$, therefore $FD=FE$. As furthermore $O_1B=O_1D$ and $O_2B=O_2E$, $\odot(BED)$ is the incircle $\tau$ of $\triangle{O_1O_2F}$. As $AG\perp BD$ and $BE\perp CG$, $G\in\odot(BED)$. Moreover, as $\sphericalangle{BDG}=\frac{\pi}{2}$, $G$ is the point opposite of $B$ on $\tau$, implying that $M=FG\cap O_1O_2$ is the touchpoint of an excircle with $O_1O_2$. Therefore, as $B$ is the touchpoint of the incircle with $O_1O_2$, it is well-known that $O_1M=O_2B$ and $O_2M=O_1B$ and hence $AM=AO_1+O_1M=AO_1+O_2C=O_2C+O_2M=MC$, as required. your solution is pretty the same as mine, Just a remark, you have to study two cases, case 1: $F$ and $G$ are from the same side of $AC$ (the one you did) case 2: $F$ and $G$ are from the opposite side of $AC$ on the second case, it more or less the same, but a bit more complicated, since $O$ become the incenter, while $G$ the antipodale of the $F$-extouch point. We're remained to prove the fact that the antipodale of the $A$-extouch point in $\triangle{ABC}$ lies on $AA_1$ where $A_1$ is the touche point betwen the incircle of $\triangle{ABC}$ and the side $BC$. This one can be proved easly by applying the homotety which maps the incenter to the $A$-excenter.