Consider the real Cartesian plane $\mathbb{R}$, with lines and betweenness but define a different notion of congruence of line segments using the distance function given by the sum of the absolute values $d(A, B) = |a_1 - b_1| + |a_2 - b_2|$ Show that the axioms $C_1 , C_2 , C_3 $ hold and draw a circle with radius 1 and center $(0,0)$
I believe that the circle of radius one at $(0,0)$ looks like a diamond i think i was also able to show $C_2$ but im not sure how to show $C_1, C_3 $
$C_1$. Given a line segment AB, and given a ray r originating at a point C, there exists a unique point D on the ray r such that $AB \cong CD$.
$C_2$. If $AB \cong CD$ and $AB \cong EF$, then $CD \cong EF$. Every line segment is congruent to itself.
$C_3$. (Addition). Given three points $A$, $B$, $C$ on a line satisfying $A * B * C$, and three further points D, E, F on a line satisfying $D * E * F$, if $AB \cong DE$ and $BC \cong EF$, then $AC \cong DF$.
In the following, for $X, Y \in \mathbf{R}^2$ I'm writing $Y-X$ as $XY$.
$(C_1)$ I guess that a ray originating from $C$ with direction $v\in \mathbf{R}^2$ is the set of points $C + tv$ where $t \in \mathbf{R}_+$. The point $D$ is the one that the corresponds to the (unique) solution in $t$ of the equation $$d(A, B) = |tv_x| + |tv_y|$$ where the right-hand side of the equation is $d(C, C + tv)$.
$(C_3)$ Let $s, t \in \mathbf{R}_+$ such that $C = A + t(AB)$, $F = D + s(DE)$. Then $BC = t(AB)-AB = (t-1)AB$ and $EF = s(DE)-DE = (s-1)DE$. As $BC=EF$, then $s=t$. Using the hypotheses about the distances, this implies $$d(A, C) = t d(A, B) = s d(D, E) = d(D, F)$$