# Egyptian Triangle By Paper Folding

When the vertices of a square are joined to the midpoints of the nonadjacent sides -- eight lines in all, a starry shape is formed in which one may discern the famous 3:4:5 triangle, known as the *Egyptian* or the *rope stretchers* triangle.

Thus the diagram provides an elegant way to form this triangle by paper folding.

In fact there are thirty two such triangles, eight of each type BFA, GFE, CDE, and HDA. A proof without words makes the assertion obvious for ΔBFA. The other triangles are clearly similar to the latter.

### References

- C. Alsina, R. B. Nelsen,
*Charming Proofs*, MAA, 2010, p. 122 - L. Bankoff, C. W. Trigg,
__The Ubiquitous 3:4:5 Triangle__,*Math Magazine*, v 47, n 2 (Mar., 1974), pp. 61-70

- An Interesting Example of Angle Trisection by Paperfolding
- Angle Trisection by Paper Folding
- Angles in Triangle Add to 180
^{o} - Broken Chord Theorem by Paper Folding
- Dividing a Segment into Equal Parts by Paper Folding
- Egyptian Triangle By Paper Folding
- Egyptian Triangle By Paper Folding II
- Egyptian Triangle By Paper Folding III
- My Logo
- Paper Folding And Cutting Sangaku
- Parabola by Paper Folding
- Radius of a Circle by Paper Folding
- Regular Pentagon Inscribed in Circle by Paper Folding
- Trigonometry by Paper Folding
- Folding Square in a Line through the Center
- Tangent of 22.5
^{o}- Proof Without Words - Regular Octagon by Paper Folding
- The Shortest Crease
- Fold Square into Equilateral Triangle
- Circle Center by Paperfolding
- Folding and Cutting a Square

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

Copyright © 1996-2018 Alexander Bogomolny

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

Copyright © 1996-2018 Alexander Bogomolny71408873