Inscribed angle "V" and whose sides pass through points "A" and "B". Draw line "OA". Angle "BOA" is a central angle; call it "θ". Lines "OV" and "OA" are both radii of the circle, so they have equal lengths. Therefore, triangle "VOA" is isosceles, so angle "BVA" (the inscribed angle) and angle "VAO" are equal; let each of them be denoted as "ψ". Angles "BOA" and "AOV" are supplementary. They add up to 180°, since line "VB" passing through "O" is a straight line. Therefore, angle "AOV" measures 180° − "θ". It is known that the three angles of a triangle add
Angle trisection and the mark on the line is labeled . This ensures that . A radius is drawn to make it obvious that line segments , , and all have equal length. Now, triangles and are isosceles, thus (by Fact 3 above) each has two equal angles. Hypothesis: Given is a straight line, and , , and all have equal length, Conclusion: angle . Proof: Clearing, , or , and the theorem is proved. Again, this construction stepped outside the framework of allowed constructions by using a marked straightedge. Thomas Hutcheson published an article in the Mathematics Teacher that used a