Killing vector field of constant length on Riemannian manifolds I would like to solve next problem

A Killing vector ﬁeld $X$ on a
  Riemannian manifold $(M, g)$ ($g$ is metric) has constant length if and
  only if every integral curve of the ﬁeld $X$ is a geodesic
  in $(M, g)$.

I found here http://arxiv.org/pdf/math/0605371.pdf (Proposition 1) solution, but I don't understand what is $L$, how we define and use that $L$, and how from that (equation in proof) follows statement. 
Or there is some alternative solutions?
My definiton of Killing vector field: Let $X$ be  vector field on a Riemannian manifold $(M,g)$ and $U$ neighbourhood of a point $p \in M$. Let $\varphi: (-\varepsilon, \varepsilon) \times U \to M$ is flow of vector field $X$. Then $X$ is Killing vector field if for every $t_0 \in (-\varepsilon, \varepsilon)$ mapping $\varphi_{t_0}:U \to M$ is isometry.

Your definition of a Killing vector field $X$ implies that
\begin{equation}
g_t(Y,Z)=g_0(Y,Z)+O(t^2)\\\lim_{t \to 0}\frac{g_t(Y,Z)-g_0(Y,Z)}{t}=0
\end{equation}
the LHS is the definition of the Lie derivative i.e.
\begin{equation}
(L_Xg)(Y,Z)=0
\end{equation} 
for any vector fields $Y$ and $Z$. So
\begin{equation}
(L_Xg)(X,Y)=0\\
Xg(X,Y)-g(L_XX,Y)-g(X,L_XY)=0\\
g(\nabla _XX,Y)+g(X,\nabla _XY)-g([X,X],Y)-g(X,[X,Y])=0\\
g(\nabla _XX,Y)+g(X,\nabla _YX)=0
\end{equation}
If the integral curve of $X$ is a geodesic, then $\nabla _XX=0$ and so
\begin{equation}
g(X,\nabla _YX)=0\\
0.5Yg(X,X)=0
\end{equation}
Then $X$ has constant length. Conversely, if $X$ has constant length then $Yg(X,X)=0$ and so $g(\nabla _XX,Y)=0$ for any vector field $Y$ and so $\nabla _XX=0$. the result.  
