Ruudu ABC tipud on vastupäeva ühendatud vastaskülgede 
keskpunktidega nii, nagu on näidatud kõrvaloleval joonisel.

Tulemusel on saadud väiksem ruut EFGH. Tõestame, et 
ruutude EFGH ja ABCD pindalade suhe on 1/3

\frac{S_{EFGH}}{S_{ABCD}}=\frac{1}{5}

n = 5 (slider from 1 to 5, step 1)
α = 0 rad (slider from 0 to 180°, step 1°)
A = (2, 2)
B = (6, 2)
C = (6, 6)
D = (2, 6)
q1, a, b, c, d = Polygon(A, B, C, D)
I = Midpoint(d)
J = Midpoint(b)
K = Midpoint(c)
L = Midpoint(a)
f = Segment(I, C)
g = Segment(A, J)
h = Segment(B, K)
i = Segment(D, L)
E = Intersect(f, h)
F = Intersect(g, h)
G = Intersect(g, i)
H = Intersect(f, i)
q2, e, f_1, g_1, h_1 = Polygon(E, F, G, H)
C' = Rotate(C, α, K)
E' = Rotate(E, α, K)
D' = Rotate(D, α, I)
D'_{1} = Rotate(D, α, I)
H' = Rotate(H, α, I)
t1, d', h', i_1 = Polygon(H', I, D')
t4, k, e', c' = Polygon(E', C', K)
A' = Rotate(A, α, L)
G' = Rotate(G, α, L)
t2, a', l, g' = Polygon(L, G', A')
F' = Rotate(F, α, J)
B' = Rotate(B, α, J)
t3, f_2, j, b' = Polygon(J, B', F)