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

Tulemusena on saadud väiksem ruut EFGH. Tõestame, et
ruutuda EFGH ja ABCD pindalade suhe on 1/5.

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

n = 5 (slider from 1 to 5, step 1)
α = 3.141592653589793 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(c)
K = Midpoint(b)
L = Midpoint(a)
f = Segment(A, K)
g = Segment(B, J)
h = Segment(C, I)
i = Segment(D, L)
E = Intersect(f, i)
F = Intersect(f, g)
G = Intersect(g, h)
H = Intersect(h, i)
q2, e, f_1, g_1, h_1 = Polygon(E, F, G, H)
A' = Rotate(A, α, L)
E' = Rotate(E, α, L)
B' = Rotate(B, α, K)
F' = Rotate(F, α, K)
C' = Rotate(C, α, J)
G' = Rotate(G, α, J)
D' = Rotate(D, α, I)
H' = Rotate(H, α, I)
t1, e', a', l = Polygon(A', L, E')
t2, f', b', k = Polygon(B', K, F')
t3, g', c', j = Polygon(C', J, G')
t4, h', d', i_1 = Polygon(D', I, H')