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The following figure was draw by taking midpoints on three sides of a regular nonagon. The angles are 20 degrees, 50 degrees, and 50 degrees.

enter image description here

It's not hard to find the angles by using trigonometry. But, I wonder if it's possible to find the angles without trigonometry.

If you succeed in finding at least one angle without trigonometry, please let me know the way.

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2 Answers 2

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Hint that doesn't need many words.      

enter image description her

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    $\begingroup$ Thank you! I don't know why I missed the similar triangles. Could you let me know about $\beta=50^\circ$? $\endgroup$
    – P.-S. Park
    Sep 16, 2018 at 2:42
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    $\begingroup$ @P.-S.Park That might be the more interesting question, indeed. I first thought it would follow by angle chasing once you know $\,\alpha\,$, but it's not all that obvious. It is easy to show that $\,\beta+\gamma=100^\circ\,$ geometrically, but there still is one relation missing. $\endgroup$
    – dxiv
    Sep 16, 2018 at 3:19
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I found an elementary solution.

Originally, two lines FJ and IC are drawn and degrees of the angle KLC is asked.

Note auxiliary lines DN and GA. Since the triangle FIC is equilateral, FP:PL = 2:1. So, the length of OL is a half of FD. So is MK. That is, OL = MK. Then, the straight line through E and P is parallel to KL and it passes the midpoint of IA. Hence the angle KLC is 50 degrees.

enter image description here

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  • $\begingroup$ Nicely done (+1). $\endgroup$
    – dxiv
    Sep 21, 2018 at 19:07

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