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How to divide an angle equally?
The problem of angle trisection is one of the three major geometric drawing problems put forward by the ancient Greeks 2400 years ago, that is, any angle is divided into three equal parts with compasses and rulers. The difficulty of the problem lies in the limitation of drawing tools. The ancient Greeks demanded that only rulers (rulers without scales, only straight lines) and compasses be used in geometric drawing. This problem attracted many people to study, but no one succeeded. In 1837, Fanzier (18 14- 1848) proved that this is an impossible proportional drawing problem by algebraic method.

In the process of studying angle trisection, special curves such as clam shell line, heart line and conic curve are found. It has also been found that as long as the subject of "drawing with a ruler" is abandoned, it is not a difficult problem to divide the angle into three parts. The ancient Greek mathematician Archimedes (287- 2 BC12) found that as long as a point was fixed on the ruler, the problem could be solved. The method is briefly introduced as follows: add a little p to the edge of the ruler, the foot tail is O, let the angle to be divided into three parts be ∠ACB, take C as the center and OP as the radius, and make a semicircle with intersecting edges at A and B; Make point O move along the CA line, point P move on the circumference, and connect OPB when the ruler passes through point B (see figure). Because OP = PC = CB, ∠ COB = ∠ ACB/3. The tools used here are not limited to rulers, and the drawing method does not conform to the postulate.

Another mechanical drawing method can be angle trisection, which is briefly introduced as follows:

As shown on the right: ABCD is a square, let AB move parallel to CD, and AD turns to DC clockwise around D. If DA arrives at DC just as AB arrives at DC, then the locus AO of their intersection, that is, the curve, is called the bisector.

Let A be any point on the AC arc, and we will bisect ADC. Let DA intersect with bisector AO at R, let parallel lines passing through R intersect with AD and BC at A and B, let t, u intersect with trisection of AD, and let parallel lines passing through T and U intersect with trisection AO at V and W, then DV and d W will bisect ADC.