Given convex quadrilateral ABCD
E, F, G, H are midpoints of AB, BC, CD, DE
M, N, P, Q are midpoints of AF, BG, CH, DE
Prove that Area(ABCD)/Area(MNPQ) = 8

Best regards,
Bui Quang Tuan

Suppose R, S, T, U are midpoints of MN, NP, PQ, QM and X is midpoint of AC.
Denote X //= Y as "X is parallel and equal Y".

Because E, M, X are midpoints of AB, AF, AC so EM //= 1/2 BF and MX //= 1/2 FC, therefore E, M, X are collinear and M is midpoint of EX (1)
Similarly G, P, X are collinear and P is midpoint of GX (2)
From (1), (2) MP //= 1/2 EG.
Moreover RS //= 1/2 MP then RS //= 1/4 EG (3)
Similarly RU //= 1/4 FH (4)

Note that RSTU and EFGH are Varignon's parallelograms of MNPQ and ABCD

Also note that area of the parallelogram can be calculated by two sides and angle between them or by two diagonals and angle between them.

Based on (3), (4), by comparing of areas of two parallelograms RSTU and EFGH, we can show that:
Area(RSTU) = 1/8 Area(EFGH)
From this, of course Area(ABCD)/Area(MNPQ) = 8

GENERALIZATION

It is interesting that this fact can be some how generalized for any convex n gon A1A2A3... An. For it we define (n,k) gon of convex n gon. (n>=4, k>=1, k<=n-1)
By one direction (for example clockwise) we connect each vertex of n gon A1A2... An with midpoint of side of next k sides from the vertex. We get n segments. By connecting all midpoints of these n segments we get one n gon and name it as (n, k) gon of given n gon A1A2A3... An.
Of course by other direction (anti clockwise) we can get another (n, k) gon of n gon A1A2A3... An. So with given n gon and given k we can construct two (n, k) gons by two directions.

With n>=5, I observer that ratio Area(n gon)/Area((n, k) gon) is not constant but:

"two (n, k) gons of given convex n gon with given k obtain the same one area value"

I still not have any idea to prove it.

Best regards,
Bui Quang Tuan