Using Euclid Elements, is it possible to bisect a line at an angle other than 90 degrees?

Question

From Euclid's Elements, Book 1, Proposition 10 shows that, the line is bisected at right angles. Is it possible to bisect a line at any angle other than 90 degree?

Answer 1

Take points $A$ and $B$ on the line $\ell$. Draw a circle at $A$ with radius $r = AB$, and let the other point of intersection be called $P$. Do the same for $B$, producing $Q$. You now have four points on a line, $P-A-B-Q$.

Draw a circle at $P$ with radius $PB$, and a circle at $B$ with the same radius. These intersect at two points, and the segment $S_1$ joining these points meets $\ell$ perpendicularly at $A$.

Do the same to produce a segment $S_2$ through $B$. Take an endpoint of $S_1$ and call it $X$; take the endpoint of $S^2$ on the other side of $\ell$, and call it $Y$. Then $XY$ is a bisector of $AB$, and the bisection is at $45$ degrees instead of $90$.

Here's a figure; I've drawn in the (green) circles used to construct $S_1$, but omitted those used to construct $S_2$, since they're very similar.

Answer 2

You can bisect a line segment with a line of almost any direction you please:

Build a triangle $ABC$ on one side; then build a congruent triangle $BAD$ on the other side with whatever method you please. Line $CD$ will bisect $AB$.

Answer 3

Let the segment be $AB$. Draw a line $L$ through A which is different from $AB$ but otherwise arbitrary. Draw a line $L'$ parallel to $L$ through $B$ using Euclid's I.31. Pick a point $C \in L$ different from $A$. Construct using Euclid's I.2 an equal segment $|BD|=|AC|$ with $D \in L'$, so that $C,D$ lie on opposite halfplanes cut by $AB$. Then $ACBD$ is a parallelogram, $AB$ and $CD$ are its diagonals, and therefore $AB \cap CD$ is the midpoint of $AB$.

Answer 4

A construction graphic may help to illustrate this concept. The line between circle centers is bisected by lines at $30, 45$ and $60$ degrees. These are the angles of the line between circle centers and the bisectors.