Aquí se puede observar el proceso para construir
un triángulo equilátero que sirva de base para el prisma.

C = (-1.588825868834104, 3.2484682982202133)
A = (-5.9383245684722334, 2.9377898196746326)
A' = (5.174405625658151, 5.208132547507722)
d = Line(A, A')
e = Line(C, d)
B = (-6.607478214570405, 7.072203418781205)
f = Line(B, d)
poly1, c, a, b = Polygon(A, B, C)
B' = Point(f)
C' = Point(e)
poly2, c', a', b' = Polygon(A', B', C')
A'' = Point(d)
p = OrthogonalLine(A'', d)
I = Intersect(e, p)
H = Intersect(f, p)
r = Circle(I, H)
q = AngularBisector(e, f)
J = Intersect(p, q)
L, M = Intersect(r, q)
s = Ray(A'', M)
t = OrthogonalLine(M, s)
B'' = Intersect(t, f)
C'' = Intersect(e, t)
poly5, n, d_2, o = Polygon(A'', C'', B'')