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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 1010.
Applying intersecting chords theorem on the circumcircle of the nonagon, CJ×DJ = AJ×EJBut EJ=AK, DJ=AG, CD=DK, thusCJ = CD+DJ = DK+DJ = DK+AGCJ×DJ = (DK+AG)×DJ = DK×DJ+AG×DJ = DK×DJ+AG²AJ×EJ = AJ×AKHence, AJ×AK = DJ×DK+AG²AJ×AK − DJ×DK = AG²
<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)=DJ^2+DJ*CD=AG^2+DJ*CD=AG^2+DJ*DK
We can show that Tr. ADJ is isoceles and Tr. KDE is equilateral and that Tr.s ADK & DEJ are congruent.Now since Tr. ACJ & DEJ are similar CJ / EJ = AJ / DJ. So ( DK + DJ) / AK = AJ / DJ. Therefore DJ. DK + DJ ^2. = AK.AJ. Hence AJ. AK - DJ. DK = DJ^2 = AD^2 = AG^2.Sumith PeirisMoratuwaSri Lanka