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Let $AC$ be a line segment in the plane and $B$ a point between $A$ and $C$. Construct isosceles triangles $PAB$ and $QBC$ on one side of the segment $AC$ such that $\angle{APB} =\angle{BQC}= 120^o$ and an isosceles triangle RAC on the otherside of AC such that $\angle{ARC} = 120^o$. Show that $PQR$ is an equilateral triangle.

I saw this question on https://artofproblemsolving.com/community/c6h57976p355627 And someone commented this:

Isn't this just Napoleon's theorem, except the triangle is degenerate (the one relating to the first fermat point?)

Others agreed with him, now I'm curious what this means.

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  • $\begingroup$ Exactly what is means: en.wikipedia.org/wiki/Napoleon%27s_theorem for a degenerate triangle $ABC$. $\endgroup$ – Jack D'Aurizio Jul 2 '18 at 21:17
  • $\begingroup$ @JackD'Aurizio but when triangle $ABC$ is degenerate, the other three triangles are supposed to be equilateral, but in this question they are not. They are just isosceles with the equal angles being $30^o$ $\endgroup$ – Shashwat Asthana Jul 2 '18 at 21:25
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    $\begingroup$ If we construct an equilateral triangle $VAB$ on the $AB$-side of $ABC$ (degenerate or not), then $\widehat{AVB}=120^\circ$ and $VA=VB$. Viceversa, if $\widehat{AVB}=120^\circ$ and $VA=VB$, then $V$ is the center of an equilateral triangle built on $AB$. $\endgroup$ – Jack D'Aurizio Jul 2 '18 at 21:28
  • $\begingroup$ @JackD'Aurizio thank you. I understood now. $\endgroup$ – Shashwat Asthana Jul 2 '18 at 21:36

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