User:Drushton2

chord lengths of circle intersections made by compass and straightedge

how to trisect an arbitrary angle using only compass and unmarked straightedge, works with acute and obtuse angles. In keeping with rules, some references of previous works are here which are inherited from several ancient cultures, such as Greece, Persia, Egypt, Japan and India. This process is reproducible by use of these tools.

Some areas of clarification I have needed to put this together are;

• 

  using the specific case of r1=r2 and centers of both circles lie on curve of the other circle, don't have to actually translate the circles, just use parallel intersections at angle intersections.

  

  with the center of the circle on the bisector, use the diameter as the center third of the intersecting parallels.

  An arbitrary angle on two dimensional surface is bisectable by drawing an initial arbitrary circle whose center is on the vertex, resulting in equal lengths of the sides of the angle,
• Lines parallel to the bisector can also be drawn using the same tools, as well as extensions of lines and scaling objects, as explained by Euclid et. al.
• When the radical axis of the circles is aligned to the angle bisector, the line containing the centers is perpendicular.
• this specific case of two circles results in total (width in first diagram, rotated to height in second diagram) of 3r, therefore centers of circles are at equal distance from sides of angles and each other.
• I haven't yet memorized Mr. Wentzel's proof, which is probably why I didn't know I couldn't do it, but it does strike me that his proof is algebraic where my proposal is not yet general but specific and geometric.
• reducible to one circle, it's diameter extended along the perpendicular to the bisector.