2

What is the best way to prove that the midpoints of BC,DE,FA are the vertices of an equilateral triangle given that ABCDEF is a hexagon inscribed in a circle with centre at O if angle AOB = angle COD = angle EOF = π/3
I could able to prove this by the help of Argand diagram using complex numbers keeping the centre of the circle at the origin but the simplification is much complicated. Is there any better approach for this purpose?

dxiv
  • 76,497
Janaka Rodrigo
  • 327
  • @peterwhy- Question must be correct because I got it from a text book and I could able to prove the result using complex numbers. I think your suggestion is impossible because it is clearly stated that ABCDEF is a hexagon. – Janaka Rodrigo Apr 23 '22 at 02:12
  • One usual method is by "finite geometric induction". The proposition is true obviously when B=C and F=A. Assume it is true for some position of A, B, C, D, E, F. It is easy to prove that it is still true if we move AB to a new position. Etc. – Apass.Jack Apr 23 '22 at 02:48
  • Content of the suggested question is very much similar to my one and it is helpful but here my intention is to find much quicker (efficient) way to prove the result. I want to know what is the best approach. – Janaka Rodrigo Apr 23 '22 at 03:17
  • @JanakaRodrigo Solving it in the complex plane is the quickest way. I somehow doubt you'll find a proof shorter than 3 lines. Which can be condensed in one line if you want to be fancy. $$ \require{cancel} \left(\cancel{\omega a} + \bcancel{b}\right) + \omega^2 \left(\bcancel{\omega b} + \xcancel{c}\right) + \omega^4\left(\xcancel{\omega c} + \cancel{a}\right) = 0 $$ – dxiv Apr 23 '22 at 03:33
  • @ dxiv - ok could you please submit your suggestion as an answer. – Janaka Rodrigo Apr 23 '22 at 03:49
  • @JanakaRodrigo I added this as an edit to my answer under the duplicate question. – dxiv Apr 23 '22 at 04:07

0 Answers0