$\textsf{\LARGE B}$

$\textsf{\LARGE C}$

E-illusion

The blue square seems to vary in size, but it's just a perception problem
that disappears when you see all of its sides (click on the blue square).

T-reality

However, in taxicab geometry, the blue square actually
varies its area according to the slope of its sides (while
both the T-length of its sides and its angles remain constant).

T metric

In the Taxicab (or Manhattan) metric, distances are measured horizontally and vertically, never diagonally.
Thus, the T-distance from B to C is the sum of the horizontal and vertical differences of its coordinates.

$\textsf{\LARGE c}$

$\textsf{\LARGE b}$

BC=c+b=|x_C - x_B|+|y_C-y_B|

T-Pythagoras

$\textsf{\LARGE a}$

Just as GeoGebra renders a segment by snapping it to the pixel grid on the screen,
we can imagine a diagonal segment made up of as small horizontal or vertical 
slices as we like: the T-distance between two points B and C will not change.

a=b+c

The square in detail

The T-perimeter of the blue square 
and the yellow square are the same, 
but the area is not:

The area of the yellow square 
is (b + c)² but the area of the 
blue square is b² + c², 
which is minimal when b = c.

a

a

a

b

c

b

c

b

c

b

c

Therefore, in taxicab geometry the T-areas coincide with the E-areas, but
the area of a T-square is NOT, in general, equal to the square of the side.

Explanation: when the 
corners are visible, 
we estimate the size 
of the square 
by its diagonal; 
when they are not, 
we value it by 
the distance between 
opposite sides 
(length of the side).

Circular compression

Because the T-length is not uniform in each direction, 
the T-circle is compressed into a square shape, 
with its diagonals parallel to the Cartesian axes.

T-Magritte

The T-distance between B and C will also be the same 
for any increasing or decreasing arc of a function.
In taxicab geometry there can be infinitely many 
minimal paths between two different points.

T-arc length

$\textsf{\LARGE a}$

I will use prefixes T and E to distinguish the Taxicab from the Euclidean metric.

T-Pythagoras (?)

All this does not simplify geometry, 
but rather complicates it. 
This is because the length of each segment 
is not uniform in direction, but depends on its slope.

$\textsf{\LARGE a}$

$\textsf{\LARGE c}$

$\textsf{\LARGE b}$

a \; ≠ \; b + c \
a^2 \; ≠  \; b^2 + c^2

B = (0, 0.25)
C = (5, 3.75)
M = Point(Segment[(x(B), y(C)), C])
L = Point(Segment[(x(M), y(B)), M])
K = Point(Segment[(x(B), y(L)), L])
J = Point(Segment[(x(K), y(B)), K])
I = Point(Segment[(x(B), y(J)), J])
H = Point(Segment[(x(I), y(B)), I])
O' = (7, 3.25)
a_2 = 4
b = 1
r: a_2 + b / 4
O = (5.75, 1.25)
polígono1, f, g, h, i = Polygon(O - (r, 0), O - (0, r), O + (r, 0), O + (0, r))
c = Circle(O, r)
S = Point(c)
radio = Segment(O, S)
P: O - (a_2 + b / 2, ((-b)) / 2)
α = 1.0716321607245183 rad (slider from 0 to 6.283185307179586, step 0.0017453292519943296)
C_1: O + (r, 0)
C' = Rotate(C_1, α, O)
C'' = Mirror(C', O)
d = LineBisector(C', C'')
Transparencia: 0
D, E = Intersect(c, d)
c1: {Polygon[P, P + (a_2, 0), P + (a_2, a_2), P + (0, a_2)]}
c2: {Polygon[P + (a_2 + b, 0), P + ((2 * a_2) + b, 0), P + ((2 * a_2) + b, a_2), P + (a_2 + b, a_2)]}
c3: {Polygon[P + (0, (-a_2) - b), P + (a_2, (-a_2) - b), P + (a_2, (-b)), P + (0, (-b))]}
c4: {Polygon[P + (a_2 + b, (-a_2) - b), P + ((2 * a_2) + b, (-a_2) - b), P + ((2 * a_2) + b, (-b)), P + (a_2 + b, (-b))]}
op5 = false
op6 = false
cc, s1, s2, s3, s4 = Polygon(C', D, C'', E)
F = Intersect(radio, polígono1)
opacidad = 1
F' = Rotate(F, 90°, O)
F'_1 = Mirror(F, O)
F'' = Mirror(F', O)
c5, f_1, f', f'_1, f'' = Polygon(F, F', F'_1, F'')
op1 = false
G = Point(Segment[(x(B), y(H)), H])
j = Segment(B, G)
k = Segment(G, H)
l = Segment(H, I)
m = Segment(I, J)
n = Segment(J, K)
p = Segment(K, L)
q = Segment(L, M)
s = Segment(M, C)
N: (x(C), y(B))
t = Segment(B, N)
e = Segment(N, C)
op2 = false
g_1 = Segment(B, C)
op7 = false
A = Point(i)
A' = Rotate(A, 90°, O)
A'_1 = Mirror(A, O)
A'' = Mirror(A', O)
c6, a, a'_2, a'_1, a'' = Polygon(A, A', A'_1, A'')
Q: (x(A''), y(A))
R: (x(A''), y(A'_1))
Q' = Mirror(Q, O)
R' = Mirror(R, O)
h_1 = Segment(A', Q')
i_1 = Segment(Q', A'_1)
j_1 = Segment(A'_1, R)
k_1 = Segment(R, A'')
l_1 = Segment(A'', Q)
m_1 = Segment(Q, A)
n_1 = Segment(A, R')
p_1 = Segment(R', A')
op8 = false
q_1 = Segment(O, F)
c_1 = Arc(c, O + (0, r), O - (r, 0))
d_1 = Arc(c, O - (r, 0), O - (0, r))
e_1 = Arc(c, O - (0, r), O + (r, 0))
r_1 = Arc(c, O + (r, 0), O + (0, r))
u = Vector(O + (sqrt(2) / 2 * (r, r)), O + (0.5 * (r, r)))
v = Vector(O - (sqrt(2) / 2 * (r, r)), O - (0.5 * (r, r)))
w = Vector(O + (sqrt(2) / 2 * ((-r), r)), O + (0.5 * ((-r), r)))
u_1 = Vector(O - (sqrt(2) / 2 * ((-r), r)), O - (0.5 * ((-r), r)))
polígono1', s_1, t_1, a_1, b_1 = Polygon(O' - (r, 0), O' - (0, r), O' + (r, 0), O' + (0, r))
op0 = true
polígono1'', f_2, g_2, h_2, i_2 = Polygon(O' - (r, 0), O' - (0, r), O' + (r, 0), O' + (0, r))
T = (-0.25, -3.25)
U = (14.42, -3.25)
ii = Segment(O + (0, r), O - (r, 0))
hh = Segment(O + (r, 0), O + (0, r))
ff = Segment(O - (r, 0), O - (0, r))
gg = Segment(O - (0, r), O + (r, 0))
op3 = false
V = (1.5, 1.5)
W = (2.5, 2.75)
Z = (4, 3)
A_1 = (1, 0.75)
B_1 = (1.75, 2.5)
o = Spline({B, A_1, V, B_1, W, Z, C}, 3)
D_1 = (-0.75, -0.5)
E_1 = (6.25, -0.5)
F_1 = (6.25, 4.75)
G_1 = (-0.75, 4.75)
c7, j_2, k_2, l_2, m_2 = Polygon(D_1, E_1, F_1, G_1)
l1: {Polygon[Q, R', 4]}
op4 = false
a' = Segment(B, C)
c_2 = Semicircle(C, B)
N' = Point(c_2)
c' = Segment(B, N')
b' = Segment(C, N')
H_1 = (-0.25, -3.5)
I_1 = (14.804545454545448, -3.5)
J_1: (-4, -4.5)
K_1: (0, -4.5)