in triangle $ABC$, $G$ is the centroid and $L$ is an arbitary line through the centroid. $K,I,J$ are the foot of the perpendiculars from $A, B, C$
i need to prove that $KA+JC=BI$
how to use the midpoints of the sides and is there any way to use the fact that centroid cut the median in 2:1 ratio
On
In any Cartesian coordinatisation of the plane, the centroid of a point set is simply the arithmetic mean. In particular, if $G$ is the origin and $L$ the $x$-axis, $0$ must be the arithmetic mean of the $y$-coordinates of $A,B,C$, i.e. with signed distances $IB+KA+JC=0$. If $B$ is alone on one side of $L$ then with unsigned distances $IB=KA+JC$.
Alternatively, one can prove the statement using the facts you mention. Introduce the midpoint $M$ of $AB$ and its ortogonal projection $N$ onto the line $L$. Since $ABJK$ is a trapezoid $(AK\parallel BJ\perp L)$, it is well known that its midline $MN$ is not only parallel to $AK, BJ$ but also satisfies $2MN=AK+BJ$.
Consider now the similar triangles $\triangle MGN\sim CGI$, and deduce, due to the fact that $CG:GM=2:1$, that $$\frac{CI}{MN}=2\iff CI=2MN=AK+BJ$$