Let $\triangle ABC$ be acute, $E,F,G$ are 3 points outside it such that $\triangle EAB$ and $\triangle BCF$ are isoscels with $$\angle AEB= \angle BFC=120^{\circ}$$ And $\triangle AGC$ is equilateral. Prove that $GB\perp EF$.
The above diagram is not totally accurate but it’s good enough. I’m searching for a cyclic quadrilateral that could help solving the problem, and I think that it should be $BPRF$, but I don’t know how to prove it.
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First, recall Napoleon's theorem. If we construct equilateral triangles $AGC$, $CHB$, and $BIA$ on the outside of triangle $ABC$, then $BG, AH, CI$
Now, observe that triangle $BEF$, rotated about $B$ by $30^\circ$ and expanded by $ \sqrt{3}$ gives us triangle $BAH$. Hence, $ AH = \sqrt{3} EF$ and is obtained by a rotation of $30^\circ$.
From above, $BG$ is a rotation of $AH$ by $60^\circ$, hence $EF \perp BG$ (and $\sqrt{3} EF = BG$).
Let $a, b, c, d, e, f, g$ the affixes of $A, B, C, E, F, G$ and : $$j = e^{\frac{2 i \pi}{3}} = -\dfrac{1}{2} + i \dfrac{\sqrt{3}}{2}$$ Notice that : $j^3 = 1$ and $1 + j + j^2 = 0$
- The triangle $AGC$ is equilateral then :
$$\dfrac{g - a}{c - a} = e^{\frac{i\pi}{3}} = e^{-i\pi} e^{\frac{4i\pi}{3}} = -j^2$$
then $g - a = -j^2 c + j^2 a$ so $g = (1 + j^2)a - j^2 c = -j a - j^2 c$.
- The triangle $BCF$ is isoscel with angle $\dfrac{2 \pi}{3}$ then :
$$\dfrac{f - b}{f - c} = e^{\frac{2i\pi}{3}} = j^2$$
then $f - b = j f - j c$. We deduce that :
$$f = \dfrac{b - j c}{1 - j}$$
- The triangle $EAB$ is isoscel with angle $\dfrac{2 \pi}{3}$ then :
$$\dfrac{e - a}{e - b} = e^{\frac{2i\pi}{3}} = j^2$$
then $e - a = j e - j b$. We deduce that :
$$e = \dfrac{a - j b}{1 - j}$$
We conclude that : $$g - b = -j^2 c - j a - b$$ and : $$e - f = \dfrac{a - j b}{1 - j} - \dfrac{b - j c}{1 - j} = \dfrac{a - (1 + j) b + j c}{1 - j} = \dfrac{a + j^2 b + j c}{1 - j} = \dfrac{j a + b + j^2 c}{j(1 - j)}$$ then : $$\dfrac{g - b}{e - f} = -j(1 - j) = -j + j^2 = -j - 1 - j = -2j - 1 = 1 - i \sqrt{3} - 1 = - i \sqrt{3} \in \mathbb{R}$$ Finally $GB \perp EF$.