Parallelogram and Four Equilateral Triangles

What Might This Be About?

Problem

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

Parallelogram and Four Equilateral Triangles - problem

Prove that $\Delta GHI$ is equilateral.

Solution

Consider triangles $GAH$ and $GDI.$

Parallelogram and Four Equilateral Triangles - solution

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).

[an error occurred while processing this directive]

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny
[an error occurred while processing this directive]
[an error occurred while processing this directive]