< PREVIOUS PROBLEM | NEXT PROBLEM >

In a regular nonagon or enneagon ABCDEFGHI, AE and CD extended meet at J, AE and DH meet at K. Prove that AJ.AK - DJ.DK = AG^{2}

Home | Search | Geometry | Problems | All Problems | Open Problems | Visual Index | 10 Problems | Problems Art Gallery | Art | 1001-1010 | Nonagon or Enneagon | by Antonio Gutierrez

Add a solution to the problem 1010 Last updated: Sep 19, 2014