It can be easily computed trigonometrically. Denote $L=BD \cap CF$, $K=CF\cap AE$ and prove that $BL=LC=KE=KF$. Now if you denote $\angle ABC =\alpha$ and $\angle CAB =\gamma$ you can express everything needed in terms of trigonometric functions with $\alpha$ and $\gamma$. That's a boring solution, anything more interesting?

Hello! Thanks for the trigo solution, I need to learn to do trigo like that Just thought I'd point out that this problem seems quite related to a lemma that appeared in the forums before: Lemma: Let the incircle of triangle $ABC$ have centre $I$ and meet $BC, CA, AB$ at $D,E,F$, and $M$ be the midpoint of $BC$. Then $ID, EF$ and $AM$ are concurrent. (Similarly, if $I_{A}$ is the $A-$excentre and is tangent to $BC, CA, AB$ at $D_{A}, E_{A}, F_{A}$ then $I_{A}D_{A}, E_{A}F_{A}$, and $AM$ are concurrent. ) For the incircle case, refer to Virgil Nicula's proof in http://www.mathlinks.ro/Forum/viewtopic.php?t=80872. The excircle case can be done in the same way. For this problem, just extend $AD$ and $BE$ to intersect, say at $X$, and let $FC$ meet $AD$ and $BE$ at $Y$ and $Z$ respectively. The two circles are an incircle and excircle of triangle $XYZ$. Using the lemma , the median of the triangle $XYZ$ passes through $N$ and $M$. Also $FY=CZ=s-y$, so so the median also passes through the midpoint of $CF$.

Megus wrote: and prove that $BL=LC=KE=KF$. Are you sure? In my diagram it is clear that $BL\neq LC$, $KE\neq KF$!

Here is my complete solution: Invalid URL $X,Y$ are the points where internal and external common tangents meet. The parallel to $XY$ through $N$ meets $PX,PY$ at $X',Y'$. The quadrilaterals $O_{2}NX'D$ and $O_{2}NEY'$, having two opposite right angles, are cyclic, and the triangle $O_{2}DE$, two of its sides being the radii, is isosceles. Therefore $\angle O_{2}X'N = \angle O_{2}DN = \angle O_{2}EN = \angle O_{2}Y'N \Rightarrow O_{2}X'Y' \mbox{isosceles}$. $O_{2}N$, being height, is also median: $NX' = NY'$, but then K, the intersection of the line $PN$ and $XY$, is the midpoint of $XY$. It is known that the midpoint of the side of a triangle is the midpoint of the points of contact of incircle and excircle with that side. Therefore $KF = KC$. Now consider the triangles $KFN, KCM$: they have both a right angle and $\angle FKN = \angle CKM$, and $KF = KC$. Then they are congruent. $KM = KN$.

Oh-hoo...oo, my solution same as @bryan_hooi's solution.

My solution: Let $X=AD \cap BE, Y=CF \cap AD, Z=BE \cap CF$ . Let $T$ be the reflection of $F$ in $O_2$ . It's well known that $CY=ZF$ . Since $AC \parallel DT , BC \parallel ET$ , so figure $\triangle ABC \cap O_1 \cap M$ and figure $\triangle DET \cap O_2 \cap N$ are homothetic with center $X$ , hence we get $X, M, N$ are collinear . ... $( \& )$ Since $XM$ pass through the midpoint of $YZ$ (see lemma 7 ) , so combine with $( \& )$ we get $MN$ pass through the midpoint of $CF$ . Q.E.D

A little typo in question, D, E, F are on O2, not O1.

Firsly, denote $L=AD \cap BE$. Let $X$ be the second intersection of the diameter $CM$ with the circle $(O_1)$. $Y$ is analogous. Because $CF$ is the common internal tangent, $CX \perp CF$ and $YF \perp CF$,so $CX || YF$. We know that the homotethy that takes $(O_1)$ to $(O_2)$ is centered at L, and because diameters $CX$ and $YF$ are parallel, then that homotethy takes $CX$ to $YF$.This shows that lines $CY$, $XF$ and $MN$ all pas through $L$ . By intersecting $CF$ with $LA$ and $LB$ at $S$ and $T$, a well known lemma gives that $LM$ bisects $ST$, and because in the triangle $LST$ , $O_1$ and $O_2$ are the incircle and excircle,the conclusion follows.