Monday, June 22, 2009

Problem 309. Regular Nonagon, Diagonals and Side

Proposed Problem
Click the figure below to see the complete problem 309.

Problem 309. Regular Nonagon or Enneagon, Diagonals and Side.
See also:
Complete Problem 309
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 4:23 PM
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5 comments:

  1. TopJune 25, 2009 at 9:07 PM

    Extend AC to J such that CJ = AB=BC=CD etc. Now angle(ACD)=140 -20=120 and thus angle(DCJ)=60. But, CJ==CD so triangle CJD is equilateral with angle(CJD)=60=angle(AED). This makes triangles AED & AJD congruent and therefore, AE=AJ=AC+AB. QED

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  2. AnonymousMay 10, 2010 at 5:15 AM

    Correct

    ReplyDelete
  3. Sumith PeirisSeptember 5, 2015 at 3:49 AM

    Extend AB and ED to meet at X. Then Tr.s AEX and BDX are both equilateral and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Sumith PeirisSeptember 5, 2015 at 11:34 AM

    I tried Ptolemy for cyclic quads but failed so far. Anyone can throw light on this?

    ReplyDelete
  5. Geek37November 14, 2018 at 3:45 AM

    Here is my solution...

    https://www.youtube.com/watch?v=WDxrXaYtGls

    ReplyDelete

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