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

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

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

n = 1 (slider from 1 to 5, step 1)
α = 0 rad (slider from 0 to 180°, step 0.017453292519943295)
A = (2, 1)
B = (6, 1)
C = (6, 5)
D = (2, 5)
q1, a, b, c, d = Polygon(A, B, C, D)
I = Midpoint(A, B)
J = Midpoint(B, C)
K = Midpoint(C, D)
L = Midpoint(D, A)
f = Segment(A, J)
g = Segment(B, K)
h = Segment(C, L)
i = Segment(D, I)
E = Intersect(f, i)
F = Intersect(g, f)
G = Intersect(h, g)
H = Intersect(i, h)
q2, h_1, e, f_1, g_1 = Polygon(H, E, F, G)
A' = Rotate(A, α, I)
E' = Rotate(E, α, I)
B' = Rotate(B, α, J)
F' = Rotate(F, α, J)
C' = Rotate(C, α, K)
G' = Rotate(G, α, K)
D' = Rotate(D, α, L)
H' = Rotate(H, α, L)
t1, c', k, g' = Polygon(K, G', C')
t2, b', j, f' = Polygon(J, F', B')
t3, a', i_1, e' = Polygon(I, E', A')
t4, d', l, h' = Polygon(L, H', D')