Three equilateral triangles

Use complex numbers. Put $G_1=0$, $B=1$ and $BD=z$ (so $BE=ze^{i\pi/3}$). Then, letting $w=e^{i\pi/3}$, $$H=\frac13(w^2+2(1+zw))=\frac13(w^2+2zw+2)$$ $$I=\frac13(w^4+2(1+z))=\frac13(w^4+2z+2)$$ But $w^2+2=w(w^4+2)$, so $$H=w\cdot\frac13(w^4+2z+2)=wI$$ Since $|w|=1$ and $\arg w=60^\circ$, $G_1H=G_1I$ and $\angle HG_1I=60^\circ$, so $\triangle HG_1I$ is equilateral.