Uh, the idea of this problem is all the same with ISL 2009 G2 ! . Due to the equality $AM \times MB = AN \times NC$ power of $M$ and $N$ w.r.t circumcircle of $\triangle{ABC}$ is equal, so $OM^2 - R^2 = ON^2 - R^2$ so $OM = ON$ and that means the perpendicular bisector of $MN$ passes through $O$ which is fixed. So we are done.

Due to the equality $AM \times MB = AN \times NC$ power of M and N w.r.t circumcircle of $\triangle{ABC}$ is equal can anyone explain why it has the same power??

The definition of the power of a point (say $P$) w.r.t a circle with radiius $r$ is equal to $OP^2 - r^2$. If line $OP$ hits that circle and point $A$ and $B$ (assume that $O$ lies between $P$ and $B$) then $OP^2 - r^2 = (OP-r)(OP+r) = (OP-OA)(OP+OB) = -PA \cdot PB$. That means if we draw any chord $CD$ of the circle passes through $P$, we have $CP \cdot DP = PA \cdot PB$. I hope it is clearer now.

Oh it is very nice thanksss