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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.

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Construction with straightlines is preserved during the affine transformation. However the bisector is not preserved, otherwise bisector in triangles would coincide with medians.

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    $\begingroup$ How can you prove that constructions with straight lines are preserved upon affine transformations? In fact, they preserve parallel lines and ratios in a line, but... $\endgroup$
    – Dr. Mathva
    Mar 24, 2020 at 18:44
  • $\begingroup$ 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 $f(P(a,b))=P(f(a),f(b))$ $\endgroup$ Mar 24, 2020 at 18:49
  • $\begingroup$ That is exactly what I would like to prove... but don't know how. Could you please elaborate further? $\endgroup$
    – Dr. Mathva
    Mar 25, 2020 at 10:28

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