Prove that if a real number, a, is constructible, then the real number (a/3) is also constructible.
I am not really sure how you want a solution. I can show this by actually doing a construction. But the theorem says, let $\displaystyle \mathbb{F}$ be the field of all constructable real numbers. Then it is an extension field over the rationals, $\displaystyle \mathbb{Q}<\mathbb{F}<\mathbb{R}$.
Thus, the construcable field contains the rationals. Now since it is a field it is closed hence $\displaystyle a/3\in \mathbb{F}$
There is nothing quite like giving the construction to prove something is
constructible
Draw a line, mark a segment of length a on the line.
Draw a line not parallel to the original line through one end of the segment.
Set the compass to some non-zero radius and mark a point at this radius
from the point of intersection of the two lines on the second line. Now
mark a second point at this radius from the previous point, and again mark a
third point at this radius from the second.
Draw a line from the third point to the other end of the original segment.
Now draw a line parallel to this last line through the first point on the
second line. then the segment from the point where the first two lines
intersect to the point at which the parallel meets the first line is a segment
of length a/3.
(You should know how to draw a line parallel to a given line through any
point not on the given line).
RonL