Derivation of the Geodesic Equation From page 40 of A. Schild and J. L. Synge's "Tensor Calculus", I'm having issues understanding the following mathematical steps ( I feel like it's simple algebra that I'm messing up. We have, $a_{rm}$ with varying Latin indices being the metric tensor and $p^{r} = \frac{dx^{r}}{dt}$ is the derivative of the coordinates w.r.t. the parameter $t$.

$$a_{rm}\frac{dp^{m}}{dt}+\frac{ \partial a_{rm}}{\partial
 x^{n}}p^{m}p^{n}-\frac{1}{2}\frac{\partial a_{mn}}{\partial
 x^{r}}p^{m}p^{n}=0  $$ By a mere rearrangement of dummy suffices, we
   have identically
$$ \frac{ \partial a_{rm}}{\partial x^{n}}p^{m}p^{n} =
 \frac{1}{2}\left(\frac{ \partial a_{rm}}{\partial x^{n}}+\frac{ \partial
 a_{rn}}{\partial x^{m}}\right)p^{m}p^{n} $$

Probably going to get down voted for this, but I'd rather not press on confused.
 A: Substituting line two into the first line, we have 
$$ a_{rm}\frac{dp^{m}}{dt}+\frac{1}{2}\left(\frac{ \partial a_{rm}}{\partial x^{n}}+\frac{ \partial
 a_{rn}}{\partial x^{m}}\right)p^{m}p^{n}-\frac{1}{2}\frac{\partial a_{mn}}{\partial x^{r}}p^{m}p^{n}=0 
$$
After a little bit of factoring and rewriting the first term we get:
$$ a_{rm}\frac{d^{2}x^{m}}{dt^{2}} + \frac{1}{2}\left(\frac{ \partial a_{rm}}{\partial x^{n}}+\frac{ \partial
 a_{rn}}{\partial x^{m}} - \frac{\partial a_{mn}}{\partial x^{r}} \right)\frac{dx^{m}}{dt}\frac{dx^{n}}{dt} $$
Noticing the bracketed term with the factor of one half is the Christoffel symbol of the first kind:
$$ a_{rm}\frac{d^{2}x^{m}}{dt^{2}}+[mn,r]\frac{dx^{m}}{dt}\frac{dx^{n}}{dt}=0  $$
Multiplying through by $ a^{rs} $:
$$ \delta^{s}_{m}\frac{d^{2}x^{m}}{dt^{2}} + a^{rs}[mn,r]\frac{dx^{m}}{dt}\frac{dx^{n}}{dt}=0 $$
In the second term we have the Christoffel symbol of the second kind, $\Gamma^{s}_{mn} $ which leads us to: 
$$ \frac{d^{2}x^{s}}{dt^{2}} + \Gamma^{s}_{mn}\frac{dx^{m}}{dt}\frac{dx^{n}}{dt}=0 $$
the Geodesic Equation!
