tag:blogger.com,1999:blog-6933544261975483399.post7327917230349458201..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1010: Regular Nonagon or Enneagon, Diagonals, Metric RelationsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-63894909761751723452015-09-05T22:50:39.325-07:002015-09-05T22:50:39.325-07:00We can show that Tr. ADJ is isoceles and Tr. KDE i...We can show that Tr. ADJ is isoceles and Tr. KDE is equilateral and that Tr.s ADK & DEJ are congruent.<br /><br />Now since Tr. ACJ & DEJ are similar CJ / EJ = AJ / DJ. So ( DK + DJ) / AK = AJ / DJ. Therefore DJ. DK + DJ ^2. = AK.AJ. <br /><br />Hence AJ. AK - DJ. DK = DJ^2 = AD^2 = AG^2.<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-53071698899020006562014-05-05T17:26:40.636-07:002014-05-05T17:26:40.636-07:00<KED=<KDE=1/2*3/9*360=60 so DEK is equilater...<KED=<KDE=1/2*3/9*360=60 so DEK is equilateral. <DAE=1/2*1/9*360=20 and <CDA=1/2*2/9*360=40 so ADJ is isosceles. AGD is also equilateral. Then AJ*AK=EJ*AJ=DJ*CJ=DJ(DJ+CD)<br />=DJ^2+DJ*CD=AG^2+DJ*CD=AG^2+DJ*DKIvan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-84922587450139160562014-05-03T18:25:58.275-07:002014-05-03T18:25:58.275-07:00Applying intersecting chords theorem on the circum...Applying intersecting chords theorem on the circumcircle of the nonagon, <br />CJ×DJ = AJ×EJ<br /><br />But EJ=AK, DJ=AG, CD=DK, thus<br />CJ = CD+DJ = DK+DJ = DK+AG<br /><br />CJ×DJ = (DK+AG)×DJ = DK×DJ+AG×DJ = DK×DJ+AG²<br />AJ×EJ = AJ×AK<br /><br />Hence, <br />AJ×AK = DJ×DK+AG²<br />AJ×AK − DJ×DK = AG²Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com