Is the triangle equilateral? (resolve without using trigonometry)

Question

I have been asked the following question. Take a circle, and three arcs of magnitude $\dfrac{\pi}{3}$ (or you could say the corresponding centered angles are of $60$ degrees) each, that do not intersect with each other, say arc $AB$, arc $CD$ and arc $EF$.

For each of the three remaining arcs take the corresponding chords $BC$, $DE$ and $FA$. Let $M, N, P$ respectively be the middles of these three line segments. Is the triangle $MNP$ equilateral?

The person that asked me this question (a mathematician) did not know if the statement is true. I asked him where did he find it and he told me that his barber asked him. I tried to solve it using Euclidean geometry and I failed. I found a positive solution using trigonometry (I may present it here later). I tried repeatedly to solve it again only using Euclidean geometry and I failed. Any idea? (I mean no analytic geometry or complex numbers, just to prove $MN=NP=PM$ using Euclidean geometry)

Edit : I will present the solution that I have, that is based on trigonometry. Let $O$ be the center of the circle, $R$ its radius and a,b,c respectively the magnitudes of the angles BOC, DOE and FOA respectively. Then $OM=R\cos\frac{a}{2}$, $ON=R\cos\frac{b}{2}$, $OP=R\cos\frac{c}{2}$.

The angle $MON$ is equal to $MON=\frac{a}{2}+\frac{\pi}{3}+\frac{b}{2}$ and taking into account that $a+b+c=\pi$ we get that $MON=\frac{5\pi}{6}-\frac{c}{2}$.

Thus $\cos(MON)=\cos(\frac{5\pi}{6}-\frac{c}{2})= \cos\frac{5\pi}{6}\cos\frac{c}{2}+\sin\frac{5\pi}{6}\sin\frac{c}{2}$ from which we deduce that $$ \cos(MON)= -\frac{\sqrt{3}}{2}\cos\frac{c}{2}+\frac{1}{2}\sin\frac{c}{2}$$

Since $\frac{c}{2}=\frac{\pi}{2}-(\frac{a}{2}+\frac{b}{2})$ we have that $\sin\frac{c}{2}=cos(\frac{a}{2}+\frac{b}{2}) =\cos\frac{a}{2}\cos\frac{b}{2}-\sin\frac{a}{2}\sin\frac{b}{2}$. We get $$ \sin\frac{c}{2}-\cos\frac{a}{2}\cos\frac{b}{2}=-\sin\frac{a}{2}\sin\frac{b}{2}. $$

Apply now the law of cosines in the triangle $MON$, in order to compute $MN^2$. Our main concern is to find a formula that is symmetric to any permutation of $a,b,c$, which will imlpy that the same formula is valid for $NP^2$ and $PM^2$.

$$ MN^2 = OM^2+ON^2-2 OM\cdot ON \cos(MON) \\ = R^2 \cos^2\frac{a}{2}+ R^2 \cos^2\frac{b}{2} -2 R\cos \frac{a}{2}R\cos\frac{b}{2} \cos(MON)\\ =R^2\Big(\cos^2\frac{a}{2}+ \cos^2\frac{b}{2} -2 \cos \frac{a}{2} \cos\frac{b}{2} \cos(MON)\Big)\\ =R^2\Big[\cos^2\frac{a}{2}+ \cos^2\frac{b}{2} -2 \cos \frac{a}{2} \cos\frac{b}{2}\big (-\frac{\sqrt{3}}{2}\cos\frac{c}{2}+\frac{1}{2}\sin\frac{c}{2}\big) \Big]\\ = R^2 \Big( \sqrt{3} \cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2} +\cos^2\frac{a}{2} + \cos^2\frac{b}{2} -\cos \frac{a}{2}\cos\frac{b}{2}\sin\frac{c}{2}\Big)\\ =R^2 \Big( \sqrt{3} \cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2} +\cos^2\frac{a}{2} + \cos^2\frac{b}{2} +\cos^2\frac{c}{2}-1 +\sin^2\frac{c}{2}- \cos \frac{a}{2}\cos\frac{b}{2}\sin\frac{c}{2}\Big)\\ =R^2[-1+\sqrt{3} \cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2} +\cos^2\frac{a}{2} + \cos^2\frac{b}{2} +\cos^2\frac{c}{2}\\ \qquad\qquad\qquad\qquad\qquad\qquad +\sin\frac{c}{2}(\sin\frac{c}{2}-\cos \frac{a}{2}\cos\frac{b}{2})\Big]\\ =R^2\Big[-1+\sqrt{3} \cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2} +\cos^2\frac{a}{2} + \cos^2\frac{b}{2} +\cos^2\frac{c}{2} -\sin\frac{a}{2}\sin\frac{b}{2}\sin\frac{c}{2}\Big] $$

It seems to me impossible to translate this trigonometric proof to a pure euclidean geometry proof. This is the reason that I made the post, for someone to suggest a pure proof based on euclidean geometry.

Edit3: As pointed out by @Blue the formula is simlpified to $$MN^2= R^2(1+\sqrt{3}\cos\frac{a}{2} \cos\frac{b}{2}\cos \frac{c}{2}+ \sin \frac{a}{2} \sin \frac{b}{2} \sin \frac{c}{2}).$$

Answer 1

It is not difficile but something tedious solve this problem without trigonometry. Let a circle of equation $x^2+y^2=r^2$ and put $A=(x_A,y_A),\cdots,F=(x_F,y_F)$. The triangles $\triangle{OAB},\triangle{OCD}$ and $\triangle{OEF}$ are clearly equilateral so we have $$\overline{AB}^2=r^2=(x_A-x_B)^2+(y_A-y_B)^2\Rightarrow 2(x_Ax_B+y_Ay_B)=r^2\space\space (1)$$ and similarly$$\overline{CD}^2=r^2\Rightarrow2(x_Cx_D+y_Cy_D)=r^2\space\space (2)\\\overline{EF}^2=r^2\Rightarrow2(x_Ex_F+y_Ey_F)=r^2\space\space (3)$$ Besides$$M=\left(\frac{x_B+x_C}{2},\frac{y_B+y_C}{2}\right)\\N=\left(\frac{x_D+x_E}{2},\frac{y_D+y_E}{2}\right)\\P=\left(\frac{x_A+x_F}{2},\frac{y_A+y_F}{2}\right)$$ It follows $$4\overline{PM}^2=(x_A+x_F-x_B-x_C)^2+(y_A+y_F-y_B-y_C)^2\\4\overline{PN}^2=(x_A+x_F-x_D-x_E)^2+(y_A+y_F-y_D-y_E)^2\\4\overline{NM}^2=(x_B+x_C-x_D-x_E)^2+(y_B+y_C-y_D-y_E)^2$$ We know by trigonometry that the request is true and then what we have to do is simply check the equalities $\overline{PM}=\overline{PN}=\overline{NM}$ using $(1),(2),(3)$ and the fact that $A,B,C,D,E,F$ are in the circle.(Easily we must have for example $4\overline{PM}^2-4\overline{PN}^2=0$ and $4\overline{PM}^2-4\overline{NM}^2=0$)