# On the construction of an angle bisector only with an straightedge

Given only an angle and a straightedge, I was wondering if it is possible to construct the angle bisector. Please think of the straightedge as a line/segment rather than the normal ruler.

This problem seems easier in the projective plane (no issues regarding parallel lines). The problem could somehow be related to Pappus's Theorem, although I doubt it. There is something, however, that sounds more promising: the construction of harmonic conjugates, and we can definitively deal with that using only a straightedge.

This image taken from Wikipedia is the one that made me think that harmonic conjugates could be helpful. (Moreover, as I said before, they can be constructed with a straightedge.) Of course, I accept other methods and viewpoints ;)

Notwithstanding, please note that I do not even know if this construction is possible, so you might also find some contradictions to its constructibility.

• If $f$ is affine transformation $l(A,B)$ is a function that generates line from points $A,B$ and $P(a,b)$ is a function that generates intersection point of lines $a,b$ then you want to show that $f(l(A,B))=l(f(A),f(B))$ and $f(P(a,b))=P(f(a),f(b))$ Mar 24, 2020 at 18:49