A fundamental algebraic property of the Arithmetic Mean states that the sum of squares of deviations (and hence the mean square deviation, $\frac{1}{N}\sum (x_i - A)^2$) is minimum when taken about the arithmetic mean (i.e. when $A = \bar{x}$).
Proof: Let $S = \sum (x_i - A)^2$. Differentiating with respect to $A$ and setting to zero yields $-2\sum (x_i - A) = 0 \implies A = \frac{\sum x_i}{N} = \bar{x}$.
Note: The sum of absolute deviations $\sum |x_i - A|$ is minimum when taken about the Median.