Dividing a Segment into N parts:
Similar Right Triangles
A Mathematical Droodle: What Is This About?
What if applet does not run? |
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander BogomolnyDividing a Segment into N parts:
Similar Right Triangles
Following is a simple algorithm for finding the Nth part of a given segment,
What if applet does not run? |
Let AB be a given segment and C a point on the line AB satisfying
Indeed, triangles AEC and APE are right and share an angle at A. They are therefore similar implying a proportion
(1) | AP / AE = AE / AC |
However, AC = N·AB = N·AE. Thus (1) gives
(2) | AP = AE² / AC = AE / N = AB / N. |
This is a clear generalization of the standard procedure of dividing a segment into two (N = 2).
References
- 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
- Divide Triangle by Lines Parallel to Base
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny71672635