Question
On the small arc $BC$ of the circumscribed circle of the triangle $ABC$, consider two distinct points $M$ and $N$, different from the ends of the arc. We know that the relations $MB+ MC=MA$ and $NB + NC = NA$ exist. Show that triangle $ABC$ is equilateral.
My idea
The first thing I noticed and that I think will help us is the fact that $ACMB, ACNB, BCNM$ are inscribed quadrilaterals.
From this info, you can see that I marked some congruent angles on my drawing.
Also, I thought of the Ptolomeu Theorem which is a bit related to the sum of sides we have, but didn't get any correct answer.
Hope you can help me!
Best Answer
Ptolemy's theorem for $ABMC$ says that $AB \cdot MC + AC \cdot MB = MA \cdot BC$. We know $MA = MB + MC$. Moving everything to one side, we get the equation $$(AB - BC) \cdot MC + (AC-BC) \cdot MB = 0.$$ If we do the same thing with $N$ in place of $M$, we get $$(AB - BC) \cdot NC + (AC - BC) \cdot NB = 0.$$ If $AB - BC$ were nonzero, we could rearrange these to get $$\frac{MC}{MB} = \frac{BC-AC}{AB-BC} = \frac{NC}{NB}.$$ But this is impossible: if $M, N$ appear in that order on arc $BC$, then $MC > NC$ and $MB < NB$, so $\frac{MC}{MB} > \frac{NC}{NB}$.
Therefore $AB-BC=0$, which simplifies the first equation above to $(AC-BC)\cdot MB=0$, and we conclude $AC-BC = 0$ as well. In other words, $AB = BC = AC$.
As a side note, once we know that $AB = AC = BC = s$, Ptolemy's theorem for an arbitrary point on the circle like $M$ says that $s \cdot MC + s \cdot MB = s \cdot MA$, so $MA = MB + MC$. So for every equilateral triangle, any points $M$ and $N$ on the circle will work to construct the diagram in the problem.