Let $X,Y$ be second intersections of $A A^\prime$, resp. $B B^\prime$ and $(ABMN)$. Both $BX$ and $MN$ are perpendicular to $AX$ and so they are parallel, also $MN \parallel AY$. The quadrilateral $AXBY$ has two right angles at $X,Y$ and has two parallel sides $AY,BX$, so it is a rectangle. Hence the segments $AY,MN,BX$ are symmetric wrt to the line bisector of $MN$ and we are done. ($69$th post)

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