# How to trisect an angle (with ruler and compasses only) !

From point A draw a vertical line and another line angling off to the right by angle β which you wish to trisect. At an arbitrary point B on the vertical line construct a right angle and draw the extended horizontal line BC. The distance h is the length of the hypoteneuse of the triangle BAC, namely AC.

Draw a large number of rays from A, marking their upper endpoints D such that the distance above the extended horizontal line BC is in each case 2h. The loci of the upper endpoints describe a curve known as the Conchoid of Nicomedes ( circa 200 BC).

In polar coordinates, the Conchoid of Nicomedes has r = a + b*secθ. See below :-

Forgive the inaccuracy of my hand-sketched (=not constructed) conchoid, please.

Now we have the conchoid, we can proceed with the actual trisection. I refer you to my sketch on the left.

From the end of the hypoteneuse at C raise a right angle vertical line from the horizontal side BC. Label the point where it intersects the conchoid D. Actually, this is the only tiny area where we need to construct a segment of the conchoid, we don't need the rest of it :-)

Draw a line DA connecting D on the conchoid with the origin at A. It intersects the horizontal BC at point E. DE=2h, remember.

Now angle BAE is one third of the angle BAC which we have successfully trisected :-)

The "impossible" trick? Note that I subtly deviated from Euclid's compass and straightedge only requirement. When I wrote Draw a large number of rays from A, marking their upper endpoints D such that the distance above the extended horizontal line BC is in each case 2h., I implicitly changed the straightedge into a ruler, sneaking my change past most people :(

Although my construct shown above does in fact trisect the angle β (I won't bore you with the proof, suffice that it exists), it does remain impossible to do the trisection while complying with Euclid's compass and straightedge only requirement.