To help develop geometric reasoning, I think it is helpful to first appreciate the Euclidean analog.
For arithmetic simplicity in special relativity, let's work with a velocity of (3/5)c, where \gamma=5/4.
Visit my visualization
"robphy SIMULTANEITY spacetime diagrammer for relativity (simplified from v8c) - 2023"
https://www.desmos.com/calculator/ajktes8bp5
which produced the graphics shown below.
In the Euclidean case (E=-1), this is a slope of (3/5) with respect to the vertical (anticipating the standard convention of time running vertically upwards).
The angle with the vertical is {\rm arctan}(3/5) \approx 40.84^\circ.
So, the unit vector along the sloped line has coordinates
adjacent-side \cos(\arctan(3/5))=\frac{5}{\sqrt{5^2+3^2}} \approx \cos(40.84^\circ)\approx 0.85749
opposite-side \sin(\arctan(3/5))=\frac{3}{\sqrt{5^2+3^2}} \approx \sin(40.84^\circ)\approx 0.51449
Imagine each [geodesic-]surveyor has an odometer and a long ruler
that he holds "perpendicular" to his path,
where "perpendicular" is along tangent to the circle [with center on the surveyor path].
Think
"tangent is perpendicular to radius".
Note the
symmetry that each surveyor's long ruler intercepts the other surveyor path at location 1.1662=\frac{1}{\cos(\arctan(3/5))} along that other surveyor path.
Indeed, the associated right-triangles with adjacent-side 1 and hypotenuse 1.1662 are
isometric triangles.
Thus, each surveyor can define a coordinate-system grid with lines parallel and perpendicular to his path.
For special relativity (E=+1), move the E-slider to +1. The "future circle" in the "geometry of Minkowski spacetime" ("Minkowski-circle") is a hyperbola, whose asymptotes are along sloped lines that are parallel to the spacetime-paths of light-rays.
Rather than Euclidean-angles involving the circular-functions,
it is more natural to work with "rapidities" (Minkowski-angles) involving the hyperbolic-functions.
[One can alternatively think in terms of areas of sectors... but I won't use this idea here.]
("Time runs upwards".)
So, the unit vector along the sloped line has coordinates
adjacent-side \cosh({\rm arctanh}(3/5))=\frac{5}{\sqrt{5^2-3^2}} =\frac{5}{4} =1.25=\gamma (the time-dilation factor)
opposite-side \sinh({\rm arctanh}(3/5))=\frac{3}{\sqrt{5^2-3^2}} =\frac{3}{4} = 0.75 = \frac{3}{5}\frac{5}{4} =v\gamma
Imagine each [inertial-]observer has a wristwatch and a long ruler
that he holds "Minkowski-perpendicular" to his path,
where "Minkowski-perpendicular" is along tangent to the "Minkowski-circle" [with center on the observer worldline].
Think
"tangent is perpendicular to radius".
(The long ruler Minkowski-perpendicular to his worldline indicates the notion of simultaneity for that observer.)
Note the symmetry that each observer's long ruler intercepts the other observer's worldline at wristwatch-time 0.8=\frac{4}{5}=\frac{1}{\frac{5}{4}}= \frac{1}{\cosh({\rm arctan}(3/5))} along that other observer worldline.
(This is the symmetry of time-dilation.)
Indeed, the associated Minkowski-right-triangles with adjacent-side 1 and Minkowski-hypotenuse 0.8 are
Minkowski-isometric triangles.
Thus, each observer can define a coordinate-system grid with lines parallel and Minkowski-perpendicular to his worldline.
For the case of \gamma=2, use v_2=0.866\approx \tanh( {\rm arccosh}(2))=\frac{\sqrt{3}}{2}.
It is important to note that the implications of this Minkowski-spacetime-geometry
lead to predictions that
agree spectacularly with experiment (better than the E=0 Galilean case below).
https://en.wikipedia.org/wiki/Tests_of_special_relativity
The "(E=0)" case corresponds to the Galilean-spacetime-geometry (the geometry of the ordinary position-vs-time graph in PHY 101... but with time running upwards). The "future Galilean-circle" is a horizontal line (more generally, a "hyperplane" ruled by segments corresponding to infinite-speeds). (Imagine opening up the hyperbolas (and the light-cone, a hyperbola of radius zero) so that the asymptotes correspond to an infinite signal-speed.)
There are analogous Galilean-trig functions (as defined by the mathematician I.M. Yaglom).
(Details are in my "Spacetime Trigonometry" poster
https://www.aapt.org/doorway/Posters/SalgadoPoster/SalgadoPoster.htm
as part of
https://www.aapt.org/doorway/TGRU/ ).
In the E=0 Galilean case, the tangent-lines determined by the various inertial observers in this diagram coincide. This implies
"absolute simultaneity in Galilean relativity".... and "no time-dilation",
In the scheme, E=0 Galilean-spacetime-geometry is the very special case (...not the general case).
Most of our confusion with special relativity stems from
our difficulty with letting go of our "common sense intuition" suggested by Galilean relativity.
