Dividing a Segment into N parts:
Similar Right Triangles
A Mathematical Droodle: What Is This About?

Explanation

Following is a simple algorithm for finding the N^th part of a given segment, N > 1, see [Maor, p. 96].

Let AB be a given segment and C a point on the line AB satisfying AC = N·AB. Draw two circles. One A(B) with center A and radius AB and another O(r), r = AC/2, on AC as a diameter. The circles cross at points E and D, and P is their common projection on line AB. Then AP = AB/N.

Indeed, triangles AEC and APE are right and share an angle at A. They are therefore similar implying a proportion

However, AC = N·AB = N·AE. Thus (1) gives

This is a clear generalization of the standard procedure of dividing a segment into two (N = 2).

References

1. E. Maor, The Pythagorean Theorem, Princeton University Press, 2007

• How to divide a segment into n equal parts
• Al-Nayrizi's Construction
• Besteman's Construction
• Besteman Construction II
• Dividing a Segment by Paper Folding
• Euclid's Segment Division
• The GLaD Construction
• The SaRD Construction
• Similar Right Triangles