I'm trying to prove something but unfortunately I can't.
Let $ABC$ be a triangle and $M$ a point in $[AB]$ where $d(A,M)=d(B,M)$.Let also be $N$ be a point in $[AC]$ where $d(A,N)=d(B,N)$.
Prove that $\overrightarrow{BC}=2\overrightarrow{MN}$.
Thanks!
$1^{st}$ MODE:
prove the similarity between the 2 triangles ($ABC$,$AMN$) you have, and then calculate the ratio between the sides
to be more exhaustive, hence you proved the similarity (they have the same angles) the ratio between sides is given you by hypothesis (let's call it $k \in \mathbb{R}$) so $k\overrightarrow{MN}=\overrightarrow{BC}$.
Pay attention of what you use in the numerator/denominator of the fraction to compute $k$.
$2^{nd}$ MODE:
following the advices given you in your previous question we can conclude that $$\overrightarrow{AM}-\overrightarrow{AN}=\overrightarrow{NM}$$ $$\overrightarrow{AB}-\overrightarrow{AC}=\overrightarrow{CB}$$
by hypothesis $$\overrightarrow{AB}=2\overrightarrow{AM}$$ $$\overrightarrow{AC}=2\overrightarrow{AN}$$ then the proof follows easily (substitution :) )