With a few basic metric assumptions, I think Euclid's postulates can be translated into coordinates.
- To draw a line between two points. Suppose our points are $(a,b)$ and $(c,d)$. You can just use a straight edge to determine that line.
In coordinates $(y-b)=\frac{d-b}{c-a}x$ determines the line.
Extend a line segment arbitrarily in any direction. Similar argument, only you are given slope and y intercept. Implies global application of line's equation.
Create a circle of radius r at any center translates to: consider center (a,b) and radius r the equation$(x-a)^2+(y-b)^2=r^2$
All right angles are congruent: I think this is equivalent to saying what's perpendicular remains perpendicular after rotations of the coordinate axes or translation of the center. Euclidean vectors can be used here. Might be equivalent to the product of the slopes of perpendicular lines is -1 and remains constant under said translations.
Two lines crossing a third will intersect on the side of that third line which the alternate interior angles are less than two right angles. Equivalently, given a line and a point C not on that line, a unique line can be draw through C parallel to the first.Coordinate equiv: Lines of the same slope either have all points in common or no points in common.
So consider the construction of an equilateral triangle given a line segment.
Euclidean approach: Consider Segment AB. Center a circle at A and draw a circle through B (III). Center a circle at centered at B draw through A (III). The intersections of the circles are the same distance from A or B as A and B are to each other. Call one of the two intersections C. Draw AC and CB (I used twice). ACB is a triangle with all sides equal. Call the other point of intersection D.
Theorem: Line CD intersects AB at its midpoint.
Proof: Call the point where CD intersects AB, M. Consider triangles CDA and CDB. CD is congruent to itself. CA, CB, DA, and DB are congruent by construction. This tells us that CDA and CDB are congruent triangles by SSS. This allows application of SAS to triangles CMA and CMB, proving that AM congruent to BM.
In coordinates:
Point A=(a,b) point B=(c,d)
M=( (a+b)/2, (c+d)/2)
More throughly:
$(x-a)^2+(y-b)^2=(x-c)^2+(y-d)^2=(a-c)^2+(b-d)^2$
$x^2-2ax+a^2+y^2-2by+b^2=x^2+c^2-2cx+y^2+d^2-2dy$
$-2ax+a^2-2by+b^2-c^2+2cx-d^2+2dy=0$
$x(-2a+2c)+y(2d-2b)+a^2-c^2=0$
This is an equation of a straight line.
$x^2-2ax+a^2+y^2-2by+b^2 =a^2+c^2-2ac+b^2+d^2-2bd$
$(x-a)^2+(y-b)^2= (a-c)^2+(b-d)^2$
is the equation of a circle.
Plug in the equation for the line into the equation for the circle to solve for points C and D. Use coordinates and the pythagorean theorem to check the lengths work to make the equilateral triangles.
This changes in non-euclidean geometries.
$\vec{r_1}=\sin\phi_1\cos\theta_1 \hat{i} + \sin \phi_1 \sin \theta_1 \hat{j}+ \cos{\phi_1} \hat{k}$
$\vec{r_2}=\sin\phi_2\cos\theta_2 \hat{i} + \sin \phi_2 \sin \theta_2 \hat{j}+ \cos{\phi_2} \hat{k}$
$\cos \alpha = \sin \phi_1 \sin\phi_2 \cos \theta_1\cos\theta_2+\sin \phi_1 \sin \phi_2 \sin \theta_1 \sin \theta_2 + \cos \phi_1\cos \phi_2$
$\cos \alpha = \sin \phi_1\sin \phi_2 \cos (\theta_1-\theta_2) + \cos \phi_1 \cos \phi_2$ Where $\alpha$ is the distance between two points on the unit sphere, $\theta$s are longitude and $\phi$s are the complementary angle of the latitude. .
Here you already see a breakdown in the circle formula for geometry on the surface of a sphere.