Problem 1209

a) An inscribed trapezoid is isosceles, so if the given hexagon is inscribed, then its diagonals are equal. Suppose now that the diagonals of the hexagon $ABCDEF$ are equal. Then, for example, $ABDE$ is an isosceles trapezoid, and the line connecting the midpoints of its bases $AB$ and $ED$ is the bisector of the angle between the lines $AD$ and $BE$. Therefore, the lines connecting the midpoints of opposite sides of the hexagon $ABCDEF$ intersect at one point $O$ - the point of intersection of the angle bisectors of the triangle formed by the diagonals $AD$, $BE$, and $CF$ (if the diagonals intersect at one point, then $O$ is precisely this point).

b) In the case of a non-convex hexagon $ABCDEF$, the only significant difference is that now the line connecting the midpoints of the sides $AB$ and $ED$ can be not only the bisector of the internal angle formed by the diagonals $AD$, $BE$, and $CF$, but also the bisector of the external angle. And three angle bisectors of a triangle, among which there are both internal and external, do not always intersect at one point (the number of external bisectors must be even). Therefore, it is additionally necessary to prove that in the considered situation, the three bisectors always intersect at one point. For this, we will use the fact that the considered bisectors $l_1$, $l_2$, and $l_3$ can be numbered such that the composition of symmetries $(S_{l_1} \circ S_{l_2} \circ S_{l_3})^2$ leaves the point $A$ in place: $A \rightarrow B \rightarrow C \rightarrow D \rightarrow E \rightarrow F \rightarrow A$. Indeed, according to problem 17.37, the transformation $S_{l_1} \circ S_{l_2} \circ S_{l_3}$ is a glide reflection, and according to problem 17.22, this transformation is a symmetry if and only if the lines $l_1$, $l_2$, and $l_3$ intersect at one point.