Three equilateral triangles form a hexagon As I posted yesterday, I was learning about vectors yesterday. I know how to add and subtract them, but I can’t multiply yet. So here is an extra problem from my teacher I need help with:

Given a hexagon $ABCDEF$ and a point $P$ on the plane sarisfying that the triangles $ABP, CDP, EFP$ are equilateral.

Prove that the midpoints of segments $AF, BC, DE$ form an equilateral triangle!
To rotate a vector around any point in the plane it's the same that to rotate this vector around his tail.

Let $R^{\alpha}$ be rotation on the plane by angle $\alpha$.

Also, let $K$, $L$ and $M$ be midpoints of $AF$, $BC$ and $DE$ respectively.

Thus, $$R^{60^{\circ}}\left(\vec{KM}\right)=R^{60^{\circ}}\left(\frac{1}{2}\left(\vec{AD}+\vec{FE}\right)\right)=\frac{1}{2}R^{60^{\circ}}\left(\vec{AP}+\vec{PD}+\vec{FE}\right)=$$ 
$$=\frac{1}{2}R^{60^{\circ}}\left(\vec{AB}+\vec{PC}+\vec{FP}\right)=\frac{1}{2}R^{60^{\circ}}\left(\vec{AB}+\vec{FC}\right)=\vec{KL}$$
and we are done!
This extra problem? It's built for the tools you don't have yet.

Hints: Put the origin at $P$. For your choice of vector systems, use complex numbers. Rotation by $60^\circ$ or $120^\circ$ (which is given by multiplication by a particular complex number) will feature prominently.

The problem also makes assumptions about the configuration; it's possible to build a hexagon for which it fails. A picture of such a hexagon is included below; the blue segments from the six labeled vertices intersect at $P$, and the triangle that we hope to be equilateral is drawn in red. Clearly, it isn't equilateral in this case.



You might object - that hexagon isn't convex. True, it isn't - but neither is $ABDCFE$ using the same vertices, and that one (with the midpoints of $BD$, $CF$, and $AE$) works.