Parallelogram and Four Equilateral Triangles

Problem

$ABCD$ a parallelogram, $ABE,$ $CDF,$ and $DGA$ are similarly oriented equilateral triangles; $H$ is the center of $\Delta ABE,$ $I$ that of $\Delta CDF.$

Prove that $\Delta GHI$ is equilateral.

Solution

Consider triangles $GAH$ and $GDI.$

Certainly, $GA=GD$ and $AH=DI.$ Also, let $\angle ABC=\alpha.$ Then

\begin{align} \angle GAH &=360^{\circ}-(180^{\circ}-\alpha)-60^{\circ}-30^{\circ}=90^{\circ}+\alpha,\\ \angle GDI &=\alpha +60^{\circ}+30^{\circ}=90^{\circ}+\alpha. \end{align}

So that the two angles, and, therefore, the two triangles are equal. In particular, $\angle AGH=\angle DGI,$ implying

$\angle HGI = \angle AGD +\angle AGH -\angle DGI=\angle AGD=60^{\circ}.$

Since $GH=GI,$ $\Delta GHI$ is isosceles with the apex angle of $60^{\circ},$ so it in fact is equilateral.

Acknowledgment

The problem has been posted by Dao Thanh Oai (Vietnam) at the Short Mathematical Idea facebook page in the name of Miguel Ochoa Sanchez (Peru).

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