1. Draw a circle with any point on the edge of the circle as the center and the center as the radius. Two circles intersect to get three points. Then, draw two circles with two new points as the center and the center as the radius, and get five points on the edge of the original circle. Then draw a circle with two new points and get seven points on the edge of the original circle. Then draw a circle with the new points. In this way, the circle is divided into four parts. 2. Draw a circle with a compass, and then draw an arc with a long radius at any point on the circle, so that there will be six intersections on the circle (counting the six points at the starting point * * *), numbered as 1-6 respectively, and then connect 1-4, 2-5 and 3- in turn. That is, the center lines of 2-4 and 1-6. In this way, you keep these two centerlines and get rid of the others, and you're done. 3. Divide the circle into four parts with compasses: (with center and no ruler) 1. You can find the bisector of the circle first, which is very easy (I believe everyone will). 2. Two of them are the two endpoints of the diameter. They intersect at c, d (the chord between two of the six points in1). 4. Through simple calculation, we can know that DO is a quarter of the chord length, which is the way to break the theory! Twenty years ago, a high school graduate from China proved that anything that can be done with a ruler can be done with a raw embroidered compass. His name is Hou Xiaorong and he is now a professor at Ningbo University. Everything that can be made with a ruler can be made with only one embroidered compass, which is completely correct. In fact, you can do it twice with the root sign of the radius. 4. Napoleon's method to solve the problem of "dividing a circle with compasses"/forum/upload 2/213312.jpg Practice: take any point A of the known circle O and take A as a point. Take A and D as the center, AC and BD as the radius, respectively, and make a circle intersect G, take A as the center, OG as the radius, and intersect M and N ⊙ O, then A, M, D and N are equal to the circumference of ⊙ O. ..