Napoleon on Hinges in GeoGebra

There are great many proofs of this popular theorem. The one that the applet above was supposed to suggest is "by a hinged dissection."

See what happens when, say, triangles $GCH$ and $IAB$ are rotated - one around $G,$ the other around $I$ - until they touch.

The triangles $GCH$ and $IAB$ when rotated - one around $G,$ the other around $I$ - so $CG$ falls on $BG$ (as they should because $BG=CG,$ by the construction, and $AI$ on $BI,$ for a similar reason, touch along the common side $BH'.$

This is because all involved angles are equal $120^{\circ}:$

$\angle BGC = \angle CHA = \angle AIB=120^{\circ}.$

Now consider $\Delta GH'I.$ Its sides are those of $\Delta GHI$ so the two are equal by SSS. Additionally and importantly the angles of $\Delta GH'I$ are all $60^{\circ}.$ For example,

$ \begin{align} \angle IGH' &= \angle IGB+\angle GBH' \\ &= \angle IGB+\angle HGC \\ &= 30^{\circ}+30^{\circ}=60^{\circ}. \end{align} $

It follows that $\Delta GH'I$ is equilateral and, therefore, so is an equal to it triangle $\Delta GHI.$

This solution is a reproduction of an older page where the applet was created in Java which is does not run on mobile devices. The present page is - thanks to the GeoGebra software - contains only HTML5 and JavaScript code, both of which are supported by mobile devices.