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I am trying to understand chapter 4 in John M. Howie's book called "Fields and Galois Theory" (published by Springer). In chapter 4 regarding geometric constructions he gives the following definition:

Let $B_0$be a set of points in the plane. There are two permitted operations on the points of $B_0$:

(1) (Ruler) through any points of $B_0$, draw a straight line;

(2) (Compass) draw a circle whose center is a point in $B_0$, and whose radius is the distance between two points in $B_0$.

Any point which is an intersection of two lines, or two circles, or a line and a circle, obtained by means of the operations (1) and (2), is said to be constructed from $B_0$ in one step. Denote the set of such points $C(B_0)$, and let $B_1 = B_0 cup C(B_0)$. We can continue this process, definining
$B_n = B_{n-1} cup C(B_{n-1})$ ($n=1,2,3,...)$. A points is said to be constructible from $B_0$ if it belongs to $B_n$ for some $n$. A point that is constructible from $lbrace O,I rbrace$ is said to be constructible.

Phew, that was long. Notice, here the points $O$ and $I$ are the origon and unit, and we may choose any two points of our initial set to be these. Now, I have attempted a bisection of an angle $PQR$, which I have illustrated in the picture below.

I tried to convert this bisection to the definition given above. As I see it I need atleast three points in my initial set $B_0$ for this. For example, I can find a bisecting line with $B_0 = lbrace Q, B, A rbrace$ ($O$ and $I$ would of course be two of these points). If I, however, try with just two, I can not. For example, suppose $B_0 = lbrace Q, A rbrace$. I can construct the red circle which has point $B$ on its exterior. However, to construct $B$ it must be an intersection point of my red circle and some other circle or line, given the definition. But to construct a line passing through $B$, I need atleast one more point.

1) Is the point $C$ in the illustration therefore not constructible, if we follow the definition given by Howie?

Howie later proves that we can not square every circle. I will not state the details on how unless you want me to, but in short, his argument is that you can not construct a square with the radius of a given circle since you can not construct such a square from $B_0 = lbrace O, I rbrace$.

2) Why is this a sufficient argument? Certainly, you could square a circle given a larger $B_0$ (for the trivial case, $B_0$ could include $pi$).

I hope this was not too long to devour. Thank you very much for your patience!

Point C is obtained as the intersection of two circles, the first has center A and passes through Q, the second has center B and passes through Q. Thus the line QC is constructible.