May be following my solution for problem "Cutting Triangles off Regular Hexagon":

https://www.cut-the-knot.org/triangle/MidpointsInHexagon.shtml

is interesting.

(Please see my attached image)
Suppose, ABCDEF is regular hexagon and M, N, P, Q, R, S are midpoints of AB, BC, CD, DE, EF, FA
Of course, by similarities, Area(ABCDEF)/Area(MNPQRS) = Area(BCEF)/Area(MPQS)
In the triangle ABF:
MS = BF/2
In the trapezium ABCD:
MP = 3*BC/2
Therefore:
Area(ABCDEF)/Area(MNPQRS) = Area(BCEF)/Area(MPQS) = BF*BC/(MS*MP) = 4/3

Best regards,
Bui Quang Tuan

Time to relax. I am going to add another one.

Please see my another attached image

We can avoid using similarities because
Area(ABCDEF)/Area(BCEF) = Area(MNPQRS)/Area(MPQS) = 3/2

We can also have another solution by construction of points IJKH.

Area(MNPQRS)/Area(MPQS) = 3/2

Area(ABCDEF)=Area(IJKH)
Area(ABCDEF)/Area(MPQS) = Area(IJKH)/Area(BCEF) = 2

Area(ABCDEF)/Area(MNPQRS) = 2/(3/2) = 4/3

Best regards,
Bui Quang Tuan

The area ratio 4/3 is still true when ABCDEF is one convex hexagon with symmetric center, say O. Please see attached image!

In the triangle AEC
Area(OCE)=Area(OEF)
Area(OAC)=Area(OCD)
Area(OAE)=Area(OED)
Therefore Area(ACE)=Area(FEDC)=Area(ABCDEF)/2
Area(ABCDEF)=2*Area(ACE) (1)

In the trapezium SMNR
SM=FB/2=EC/2
RN=EC

Distance from SM to FB = 1/2 Distance from A to FB
Distance from FB to RN = 1/2 Distance from FB to EC
Therefore
Distance from SM to RN = 1/2 Distance from A to EC
and
Area(SMNR)=3/4*Area(ACE)
We also have Area(SMNR)=Area(MNPQRS)/2 so
Area(MNPQRS)=2*Area(SMNR)=3/2*Area(ACE) (2)

From (1), (2) we have result:
Area(ABCDEF)/Area(MNPQRS)=4/3

Best regards,
Bui Quang Tuan