Geometry Problems, Online Education

Problem 725: Kurschak's Dodecagon, Square, Equilateral Triangle, Midpoints, 30,60 Degrees. High School, College

In the diagram below, ABCD is a square with the equilateral triangles ABE, BCF, CDG, and ADH. Prove that (1) EFGH is a square; (2) The 8 intersections (N1,2..8) of the equilateral triangles and the midpoints (M1,2,3,4) of the sides of EFGH form a regular dodecagon. 

Kurschak Dodecagon, Square, Equilateral Triangle, Midpoint

Home | Geometry | Search | Problems | Square | Equilateral Triangle | Midpoint | Polygon | Kurschak's Tile | Open Problems | All Problems | Visual Index | 721-730 | Email | Solution/comment | By Antonio Gutierrez