So to solve the heptagon puzzle, we would hope to reduce it to this case: a bite made of two equal lines, which we can then solve by letting the vertex lie on the circle centre. We want to deform the four edges of the heptagon to two equal lengths, whilst maintaining the two endpoints where it meets the circle, and also maintaining the area of the heptagon.

If I draw an line between two non-adjacent vertices, then draw a parallel line through the vertex between them, and extend the next edge of the heptagon to meet that parallel line, and then move the middle vertex to that new intersection, then the area is preserved, but the number of edges has been reduced. I can do the same thing again to reduce the cut to two edges:

Finally to make those two edges equal, I draw a line between the two fixed vertices, drop a perpendicular bisector, and find the intersection between that and the parallel line through the remining internal vertex.

Finally, centring the circle on this point relative to the heptagon and rescaling the figure so that the arc length is equal to 1, we do indeed find this to give the maximum area, which is approximately equal to 0.0942943...