If $p - a \equiv -a \pmod{p}$ then what is $\frac{p-1}{2} \equiv ? \pmod{p}$? If $p - a \equiv -a \pmod{p}$ then what is $\frac{p-1}{2} \equiv ? \pmod{p}$? Where $p$ is an odd prime.
I read in the book, they claimed:
$$p - 1 \equiv -1 \pmod{p}$$
$$p - 2 \equiv -2 \pmod{p}$$
$$p - 3 \equiv -3 \pmod{p}$$
$$ ... $$
$$\frac{p - 1}{2} \equiv -(\frac{p - 1}{2}) \pmod{p}$$
However, I realized that the last statement was wrong because if it's true then:
$$\frac{p - 1}{2} + \frac{p - 1}{2}\equiv 0 \pmod{p}$$
$$p - 1 \equiv 0 \pmod{p}$$

Any idea?   
Thanks, 

The sequence as written doesn't make sense, since the last term does not follow the pattern of the previous ones. However, you can argue that
$$p - \left(\frac{p-1}{2}\right) \equiv -\frac{p-1}{2}\pmod{p}.$$
This does follow the same pattern as the rest of the terms.
And since
$$p - \frac{p-1}{2} = \frac{2p-p+1}{2} = \frac{p+1}{2}$$
then you have that
$$\frac{p+1}{2}\equiv -\frac{p-1}{2}\pmod{p},$$
hence $\frac{p-1}{2}\equiv -\frac{p+1}{2}\pmod{p}$.
Or peharps, it went like this:
\begin{align*}
p-1 &\equiv -1\pmod{p}\\
p-2 &\equiv -2 \pmod{p}\\
&\vdots\\
\frac{p-1}{2} = p-\left(\frac{p+1}{2}\right) &\equiv -\frac{p+1}{2}\pmod{p}
\end{align*}
and you have the wrong sign in the numerator on the right hand side?
Added. So, the rest of the proof. According to the scans Sivaram has posted, this was part of the computation of $\left(\frac{p-1}{2}\right)!^2$. The idea then is that we can get $\left(\frac{p-1}{2}\right)!$ by multiplying the terms on the right hand side, and then get it again by multiplying the terms on the left hand side. That is, we look at
\begin{align*}
p-1 &\equiv -1 \pmod{p}\\
p-2 &\equiv -2 \pmod{p}\\
&\vdots\\
p-\left(\frac{p-1}{2}\right) &\equiv -\frac{p-1}{2}\pmod{p}
\end{align*}
and note that if $p\equiv 3\pmod{2}$, then we have an odd number of negative signs on the right hand side. So we have that:
$$(-1)\left(\frac{p-1}{2}\right)! = (-1)\Bigl(1\times 2\times\cdots\times \frac{p-1}{2}\Bigr) \equiv (p-1)(p-2)\cdots\left(p - \frac{p-1}{2}\right)\pmod{p}.$$
Now, $p - \frac{p-1}{2} = \frac{p+1}{2} = \frac{p-1}{2}+1$. So therefore we have that:
\begin{align*}
\left(\frac{p-1}{2}\right)!^2 &= \left(\frac{p-1}{2}\right)!\left(\frac{p-1}{2}\right)!\\
&\equiv  \left(\frac{p-1}{2}\right)!\Biggl( -(p-1)(p-2)\cdots\left(\frac{p-1}{2}+1\right)\Biggr)\pmod{p}\\
&\equiv -\left(1\times 2\times 3\times\cdots\times\left(\frac{p-1}{2}\right)\times\left(\frac{p-1}{2}+1\right)\times\cdots\times (p-1)\right)\pmod{p}\\
&\equiv -(p-1)!\pmod{p}.
\end{align*}
But we know that $(p-1)!\equiv -1\pmod{p}$ by Wilson's Theorem, so
$$\left(\frac{p-1}{2}\right)!^2 \equiv -(p-1)!\equiv -(-1) = 1\pmod{p},$$
so if we let $x = \left(\frac{p-1}{2}\right)!$, then $x^2 \equiv 1 \pmod{p}$. This means that $p|x^2-1 = (x-1)(x+1)$, so either $p|x-1$ or $p|x+1$. that is, eithe $x\equiv 1 \pmod{p}$ or $x\equiv-1\pmod{p}$; giving: 
$$\text{either }\left(\frac{p-1}{2}\right)!\equiv -1\pmod{p}\quad\text{or}\quad \left(\frac{p-1}{2}\right)!\equiv 1\pmod{p}.$$
