Possible trisection of a line with Euclidean geometry

Question

A long time ago my geometry teach said that no one has found a way to trisect an angle with Euclidean geometry. So I trisected an angle with Euclidean geometry. I think that the proof works.

Take any rectilinear angle (ABC). Pick a point D on either vertex. Draw a circle with line BD as the diameter and B as the center. Where circle B intersects BC, label it E. Construct equilateral triangle DEF with a base of DE, with F pointing away from B. Bisect lines DF and EF (these will be called G and H, respectively). Draw the lines BG and BH. Those line are trisecting the angle.

If the angle is bigger than 180°, Bisect the angle and trisect the new angles, then take the two lines closest to the bisecting line as the trisecting lines.

Answer 1

Trisection of an angle with compass and unmarked straight edge is equivalent to solving an irreducible cubic equation algebraically...that is why it has been said to be impossible except for special angles such as 90°. While I would like to encourage you, there must be a flaw in your construction...and so my encouragement is to tell you to find the flaw.