Why does this construction work? (Explanation)

Question

In the book that I am currently studying, this construction appears (see image) to show that if $a$ and $b$ are constructibles, then $ab^{-1}$ is constructible, but I cannot understand why. I have seen in Artin's algebra book that he constructs these numbers using similar triangles and I can understand it, but this with this construction I have no idea. Please help me.

Answer 1

If you denote the points with coordinates $(0,a)$, $(-b,0)$, $(0,-1)$ and $(0,0)$ as $A$, $B$, $C$ and $O$ respectively, and draw the circle containing $A$, $B$ and $C$, it intersects your $x$-axis at some point, call it $D=(d,0)$. Then the triangles $OBC$ and $OAD$ are similar, because they are both right and angle $CAD$ = angle $CBD$ (inscribed on the same chord). Therefore $OD/OA=OC/OB$ or $d=a\cdot 1/b=ab^{-1}$. Therefore you can construct the segment of length $ab^{-1}$ if only the segments of length $a$ and $b$ can be constructed.