Question: A different point $M$ is taken from the endpoints on the small arc $AC$ of the circumcircle of a triangle $ABC$ satisfying the condition $|AB|=|AC|$. where $[BM]$ intersects $[AC]$ is $E$, the interior bisector of angle $BMC$ intersects $[BC]$ with $F$, $m(\widehat{AFB}) =m(\widehat{CFE})$ equality is achieved. Show that triangle $ABC$ is an equilateral triangle.
Drawn version: image
Here's what I've tried so far to tackle the problem: I first extended [MF] and combined it with $A$, and then I applied the trigonometric answer theorem to two triangles with umbilical points $E$ and $F$, and I got nothing. I have a feeling I should make better use of the $m(\widehat{AFB}) =m(\widehat{CFE}) $information, but I'm not getting anywhere.
Since $E$ is the midpoint of $E'E''$ and $\angle A=90^\circ$, $\triangle AEE'$ is isosceles at $E$. It follows that $\angle AEE''=2 \angle EAE'$, equivalently $E$ trisects the small arc $AC$.
Best Answer
Consider figure; CI and BJ are altitudes of ANC. Using the fact that in triangle BMC the bisector of angle BMC meets the perpendicular bisector of BC at point G , so AG is diameter of circumcircle and it is also the altitude of ABC. We have:
$\angle BMC=\angle BAC$
$\angle GMC=\angle GAC$
We mirror the construction about BC, so K and G are mirror os A and H respectively. We have:
$\angle HBC=\angle HKC=\frac{\overset{\large \frown}{HC}}2$
$\angle GAC=\frac{\overset{\large \frown}{GC}}2$
$\overset{\large \frown}{HC}=\overset{\large \frown}{GC}$
$\Rightarrow \angle HBC=\angle HAC$
$ \angle HKC=\angle HAC=\angle HKB\space\space\space\space (1)$
$\angle GBA=90^o$
Because it is opposite to diameter AG, so AB is tangent to bottom circle centered at G. So we have:
$ \angle ABH=\angle BKH\space\space\space\space (2)$
Summing (1) and (2) we get:
$\angle ABH+\angle HBC=\angle HAC+\angle HAB$
OR:
$\angle ABC=\angle BAC=\angle ACB$
that is triangle is equilateral.