In the book Geometry: Euclid and beyond, the exercise 2.20 says:
Using a ruler and rusty compass, given a line $l$ and given a segment $AB$ more than one inch long, construct one of the points $C$ which the circle of center $A$ and radius $AB$ meets $l$.(Rusty compass's radius is 1 inch)
But I don't think it can be done use ruler and rusty compass. As the picture below shown, we need to get point $C$, use ruler and rusty compass, we can easily twice long $BA$ to point $B'$ and construct perpendicular $BG,B'H$, but in order to get length of $GC$, we need to solve a quadratic equation and need to use square root operation, but some results before said that we can only get the length in $\mathbf{Q}$ use ruler and rusty compass. I am very confused about that. Is this exercise solvable?
Suppose we consider a coordinate system centred at the point $A$. Now, suppose we shrink the entire figure, keeping point $A$ fixed such that the distance between $A$ and $B$ becomes equal to $1$ inch. Hence, we can say that $B$ moves to a point $B_1$ on $AB$ such that $AB_1=1$ inch. Also, suppose that in this process, $G$ shifts to $G_1$ and $H$ shifts to $H_1$. Then, join $B_1H_1$ and $G_1H_1$.
Now, with centre $A$ draw a circle of radius $AB_1$ (which is equal to $1$ inch) that meets $G_1H_1$ at say $C_1$. Then, extend $AC_1$ to meet $GH$ at $C$ which is the required point.