$G$ is the intersection of the medians of $\triangle{ABC}$, and $K$ and $L$ are points on $AB$ and $AC$, respectively, such that $G$ is a point on $KL$. The line determined by $K$ and $L$ intersect the line determined by $B$ and $C$ at $P$. If $C$ is between $B$ and $P$, \begin{equation*} \frac{1}{KG} = \frac{1}{LG} + \frac{1}{GP} . \end{equation*}
Write $KG=x$, $GL =y$ and $LP =z$. Then we have to prove $y^2+zy =2xy+xz$.
Also let $R$ halves $BC=2a$ and mark $CP=b$.
Now, if we apply Menelaus theorem twice on triangle $GRP$
first on transversal $B-K-A$ we get $${b+2a\over a}\cdot {3\over 2}\cdot {x\over x+y+z}=1$$ and second on transversal $C-L-A$ we get $${b\over a}\cdot {3\over 2}\cdot {y\over z}=1$$ and eliminating ${b\over a}$ from both, you will get a desired formula.