Perigal's Proof of the Pythagorean Theorem


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

A few words and an explanation.

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

Explanation


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Assume b > a, so that the lines parallel and perpendicular to the hypotenuse of ΔABC are drawn in the square ACBcBa through its midpoint M. The length of both segments of these lines inside the square equal the length of the hypotenuse c. Each segment is divided by M into equal parts of length c/2.

Drop a perpendicular from M onto CBc. Its foot will land in the middle of CBc. Thus a triangle will be formed with the sides parallel to those of ΔABC, but just half as big.

From here we conclude that the endpoints on the add-on lines in square ACBcBa divide its sides (each of length b) into two parts (a + b)/2 and (b - a)/2. The difference of the two

(a + b)/2 - (b - a)/2 = a,

which explains the emergence of a square of side a after rearrangement of the pieces on the hypotenuse.

Equidecomposition by Dissection

  1. Carpet With a Hole
  2. Equidecomposition of a Rectangle and a Square
  3. Equidecomposition of Two Parallelograms
  4. Equidecomposition of Two Rectangles
  5. Equidecomposition of a Triangle and a Rectangle
  6. Equidecomposition of a Triangle and a Rectangle II
  7. Perigal's Proof of the Pythagorean Theorem
  8. Two Symmetric Triangles Are Directly Equidecomposable
  9. Wallace-Bolyai-Gerwien Theorem

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

71471324