Spherical Construction Problem about using a ruler and a compass

Spherical Construction Problem about using a ruler and a compass

I've known the following theorem:
Theorem 1: On a plane, if we have both of a primitive ruler and a
primitive compass, then we can do the same construction as we can do by
using a macro-ruler or a macro-compass.
Here, a primitive ruler is a ruler by which we can only draw a line
segment $AB$ from a point $A$ to a point $B$. By using a primitive
compass, we can only draw a circle: its center is a point $A$ and it
passes through a point $B$. A macro-ruler is a ruler by which we can do
the same construction as a primitive ruler, and we can also extend a line
segment $AB$ as long as you like. By using a macro-compass, we can draw a
circle: its center is a point $A$ and its radius is the same length as a
line segment $BC$. This means that if there exist a point $A$ and a line
segment $BC$, then we can decide the radius by fitting a macro-compass to
the line segment $BC$ and can move it to the point $A$.
I'm going to show the proof for theorem 1 at the end.
By the way, I've been interested in the construction on the surface of a
sphere. Here is my question.
Question: Can we say the same thing as above on the construction on the
surface of a sphere? Imagine that you are on the surface of a sphere and
you have a primitive ruler and a primitive compass. Then, can you do the
same construction as a macro-ruler and a macro-compass? Suppose that each
tool can work even on a sphere: by using a primitive ruler, you can draw
the shortest line segment $AB$, which is actually an arc $AB$ from a point
$A$ to a point $B$ on the surface of a sphere,.
I think we can do the same construction as macro-ruler by the same
argument as above, but to be honest, I'm not sure. I would like a rigorous
proof, whichever it's possible or not.
Here is the proof for theorem 1.
Proof for theorem 1: Let $C(A,B)$ be a circle whose center is a point $A$,
and which passes through a point $B$.
First, let's prove that if we have both of a primitive ruler and a
primitive compass, then we can do the same construction as we can do by
using a macro-ruler. For a given line segment $AB$, draw $C(A,B)$ and
$C(B,A)$. Let one of intersection points of the two circles be $C$. Draw
$C(C,A)$ and let one of intersection points of $C(A,B)$ and $C(C,A)$ be
$D$. Suppose that $D$ is not $B$. Draw C(D,A) and let one of intersection
points of $C(A,B)$ and $C(D,A)$ be $E$. Suppose that $E$ is not $C$. Draw
a line segment $AE$. Now get a line segment $AE$ which is twice as long as
$AB$. Since, we can repeat the same argument infinitely as above, now the
proof is completed.
Second, let's prove that if we have a primitive compass, then we can do
the same construction as we can do by using a macro-compass. Suppose that
there exist a point $A$ and a line segment $BC$. Draw $C(A,B)$ and
$C(B,A)$. Let one of intersection points of the two circles be $D$. Draw
$C(C,D)$ and $C(D,C)$. Let either of the two intersection points of the
two circles be $X$. Draw $C(A,X)$. Then, we get that a triangle$ADX$ and a
triangle$BDC$ are congruent. Let's prove this. Obviously, we get $AD=BD,
DX=DC$. On the other hand, we get that $\angle ADX=\angle ADB-\angle
XDB=60^{\circ}-\angle XDB$ and that $\angle BDC=\angle XDC-\angle
XDB=60^{\circ}-\angle XDB$. Hence, we get $\angle ADX=\angle BDC$. So the
proof about triangles is completed. Hence, we get $AX=BC$. Now the proof
is completed.