Let $\vec{AA_1} + \vec{BB_1} + \vec{CC_1} = 0$. Prove that the triangle is equilateral. Let $A A_1, BB_1, CC_1$ be the angle bisectors   of triangle $ABC$ and $\vec{AA_1}+\vec{BB_1}+\vec{CC_1}=0.$ Prove that $ABC$ is equilateral.
I know  that  the vectors  $\vec{AA_1}$ and $\frac{\vec{AB}}{|AB|}+\frac{\vec{AC}}{|AC|}$ ( and so on) are collinear  but not find how it can help.
Any ideas? 
 A: Using trilinear or barycentric coordinates is a sensible choice, but we may go through "pure" geometry as well. Let $B''$ be such that $A'B''$ is parallel to $BB'$ and has the same length. We have $\vec{AA'}+\vec{BB'}+\vec{CC'}=0$ iff $CC'AB''$ is a parallelogram. In such a case $B'$ has to be the midpoint of $AC$, hence $BA=BC$ by the bisector theorem. Since $C',B',B''$ have to be collinear, by Thales' theorem $C'$ is the midpoint of $AB$ and $CB=CA$. Done.

A: We have, for suitable non-zero constants $u,v,w$,
$$ u\left(\frac{\vec{AB}}{|AB|}+\frac{\vec{AC}}{|AC|}\right)+ v\left(\frac{\vec{BC}}{|BC|}+\frac{\vec{BA}}{|AB|}\right)+
w\left(\frac{\vec{CA}}{|AC|}+\frac{\vec{CB}}{|BC|}\right)=0,
$$
or
$$\frac {u-v}{|AB|}\vec{AB}+\frac {v-w}{|BC|}\vec{BC}+\frac {w-u}{|AC|}\vec{CA}=0,$$
or, using $\vec{CA}=-\vec{AB}-\vec{BC}$,
$$ \left(\frac {u-v}{|AB|}-\frac {w-u}{|AC|}\right)\vec{AB}+\left(\frac {v-w}{|BC|}-\frac {w-u}{|AC|}\right)\vec{BC}=0.$$
As $\vec{AB}$, $\vec{BC}$ are linearly independent, this menas that both coefficients are $0$.
if $u>v$, then form the coefficient of $\vec{AB}$, we find $w>u$, and then from the other coefficient $v>w$, contradiction. Likewise, $u<v$ leads to a contradiction. Hence $u=v$ and we also obtain $w=u$.
Can you take it from here?
A: 
Using the so-called angle bisector theorem, one can write
(with usual notations $a,b,c$ for the side lengths $BC,CA,AB$ resp.):
$$\tag{1}\vec{AA_1}=\dfrac{b}{b+c}\vec{AC}+\dfrac{c}{b+c}\vec{AB}$$
In the same way, by cyclic permutation:
$$\vec{BB_1}=\dfrac{c}{c+a}\vec{BA}+\dfrac{a}{c+a}\vec{BC}$$
which can be written:
$$\tag{2}\vec{BB_1}=-\dfrac{c}{c+a}\vec{AB}+\dfrac{a}{c+a}\vec{AC}-\dfrac{a}{c+a}\vec{AB}$$
and finally:
$$\vec{CC_1}=\dfrac{a}{a+b}\vec{CB}+\dfrac{b}{a+b}\vec{CA}$$
which can be written:
$$\tag{3}\vec{CC_1}=\dfrac{a}{a+b}\vec{AB}-\dfrac{a}{a+b}\vec{AC}-\dfrac{b}{a+b}\vec{AC}$$
adding (1)+(2)+(3) gives
$$\tag{4}\vec{AA_1}+\vec{BB_1}+\vec{CC_1}=\alpha \vec{AB}+\beta \vec{AC}$$
with $\alpha=\frac{c}{b+c}-\frac{c}{c+a}-\frac{a}{c+a}+\frac{a}{a+b}=\frac{ac-b^2}{(a+b)(a+c)}$ and
$\beta=\frac{b}{b+c}+\frac{a}{a+c}-\frac{a}{a+b}-\frac{b}{a+b}=\frac{ab-c^2}{(b+c)(a+c)}.$
$\{\vec{AB},\vec{AC} \}$ being a basis of the plane, we will have a zero sum in relationship (4) if and only if the two numerators in the expressions of $\alpha$ and $\beta$ are zero, i.e.,
$ac=b^2$ and $ab=c^2$, whence $\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}$ from which, denoting the common value of these ratios by $k$ we obtain $a=kb, b=kc,c=ka$. Thus $k^3=1$ which is equivalent to $k=1$. Consequence: $a=b=c$.
A: If we write this equation with position vectors we get $${1\over 3}(\vec{A}+\vec{B}+\vec{C}) = {1\over 3}(\vec{A}_1+\vec{B}_1+\vec{C}_1)$$ which means that triangles $ABC$ and $A_1B_1C_1$ share the same gravity center. But this means that every angle bisector is also a median which means $\triangle AA_1B \cong \triangle AA_1C$. But this means that angles at $B$ and $C$ are the same and further as angle at $A$ and we are done.
