All internal angles of a convex octagon ABCDEFGH are equal to each other and the edges are alternatively equal:
AB=CD=EF=GH,BC=DE=FG=HA
(we call such an octagon semiregular). The diagonals AD, BE, CF, DG, EH, FA, GB and HC divide the inside of the octagon into certain parts. Consider the part containing the centre of the octagon. If that part is an octagon, then this central octagon is semiregular (this is obvious). In this case we construct similar diagonals in the central octagon and so on. If, after several steps, the central figure is not an octagon, then the process stops. Prove that if the process never stops, then the initial octagon was regular. (A. Tolpygo, Kiev) geometrycombinatoricscombinatorial geometryoctagon