We have a triangle $ABC$. On sides $AB$, $BC$, $CA$ choosen points $A_1$, $B_1$, $C_1$, different with vertices $A$, $B$, $C$. We have a triangle $ABC$. On sides $AB$, $BC$, $CA$ choosen points $A_1$, $B_1$, $C_1$, different with  vertices $A$, $B$, $C$. 
Then for a specified $k_a$, $k_b$, $k_c$ we have
$\vec{BA_1} = k_a\vec{CA_1}$
$\vec{CB_1} = k_b\vec{AB_1}$
$\vec{AC_1} = k_c\vec{BC_1}$
Then $A_1$, $B_1$, $C_1$ are on a straight line?
 A: I think the $A_1,B_1$ and $C_1$ is wrong positioned so I made correction, that is $A_1$ is on $BC$, $B_1$ is on $AC$ and $C_1$ is on $AB$. and it is a well known co-line principle. we can have some simple proof:
there will be basic principle :
 
in $\triangle ABC$,if $A_1$ is on $BC$, then it must satisfy :
$m*\vec{AC}+n*\vec{AB}=\vec{AA_1}$  and $m+n=1$ and vice verse.
look at $\triangle A_1CC_1 $ ,if $B_1$ is on $A_1C_1$, it must have:
$m*\vec{CA_1}+n*\vec{CC_1}=\vec{CB_1}$ ......1
and $m+n=1$
since $\vec{BA_1}+\vec{A_1C}=\vec{BC}$, then $ k_{a}\vec{CA_1}+\vec{A_1C}=\vec{BC}$, so we get:
$\vec{CA_1}=\dfrac{1}{k_{a}-1}\vec{BC}$......[2]
with similar work, we can get 
$\vec{CB_1}=\dfrac{k_{b}}{k_{b}-1}\vec{CA}$......[3]
$\vec{CC_1}=\vec{BC_1}+\vec{CB}$,
$\vec{BC_1}=\vec{AC_1}+\vec{BA}=\vec{BA}+k_{c}\vec{BC_1}$, so we got:
$ \vec{BC_1}=\dfrac{1}{1-k_{c}}\vec{BA} $, then $\vec{CC_1}=\dfrac{1}{1-k_{c}}\vec{BA}+\vec{CB} $ $=\dfrac{1}{1-k_{c}}(\vec{CA}-\vec{CB})+\vec{CB}$.....[4]
put [2],[4] into 1 LHS, we get
LHS=$\dfrac{m}{k_{a}-1}\vec{BC}+n\vec{CB}+\dfrac{n}{1-k_{c}}(\vec{CA}-\vec{CB})$
replace $\vec{CA}$ with [3], we have:
LHS=$(\dfrac{m}{k_{a}-1}-n+\dfrac{n}{1-k_{c}})\vec{BC}+\dfrac{n}{1-k_{c}}*\dfrac{{k_{b}}-1}{k_{b}}\vec{CB_1}$
compare to RHS, we have :
$\dfrac{m}{k_{a}-1}-n+\dfrac{n}{1-k_{c}}=0$......[5]
and 
$ \dfrac{n}{1-k_{c}}*\dfrac{{k_{b}}-1}{k_{b}}=1 $......[6]
with [5] and [6] and $m+n=1$,
we can finally get:
$(1+\dfrac{1}{k_{a}-1}+\dfrac{1}{k_{c}-1})*\dfrac{(1-k_{c})k_{b}}{k_{b}-1}=\dfrac{1}{k_{a}-1}$
that is:
$k_{a}k_{b}k_{c}=1$
