I have a chessboard with a square marked with A as in the following figure:

$$\begin{array}{|c|c|c|} \hline
8&1&2\\ \hline
7&A&3\\ \hline
6&5&4\\ \hline
\end{array}$$

The $4$-*connected neighbors squares* of A are marked with $1, 3, 5, 7$.
The $8$-*connected neighbors squares* of A are marked with $1, 2, 3, 4, 5, 6, 7, 8$.

Then I have a chessboard with two marked squares, an *origin* square and a *destination* square: a $4$-*connected path* from the origin to the destination is a sequence of squares that connects the origin to the destination and all the squares in the sequence are $4$-connected neighbors; an $8$-*connected path* from the origin to the destination is a sequence of squares that connects the origin and the destination and all the squares in the sequence are $8$-connected neighbors. The *length* of a path is the number of squares in the sequence.

The left part of the following figure shows a $4$-connected path from the origin square marked with $1$ to the destination square marked with $7$; the length of the path is $7$. The right part of the figure shows an $8$-connected path from the origin $1$ to the destination $5$; the length of the path is $5$.

$$\begin{array}{}
\begin{array}{|c|c|c|} \hline
&6&7\\ \hline
&5&\\ \hline
&4&\\ \hline
&3&\\ \hline
1&2&\\ \hline
\end{array}&\qquad&\begin{array}{|c|c|c|} \hline
&&5\\ \hline
&4&\\ \hline
&3&\\ \hline
&2&\\ \hline
1&&\\ \hline
\end{array}
\end{array}$$

I have read in [stackoverflow](https://stackoverflow.com/questions/19984784/measure-noise-in-a-binary-bitmap-line/20081295#20081295) that the length of the shortest $4$-connected path between origin and destination is the [Manhattan distance](http://en.wikipedia.org/wiki/Manhattan_distance) between origin and destination and that the length of the shortest $8$-connected path is the [Chebyshev distance](http://en.wikipedia.org/wiki/Chebyshev_distance) between origin and destination.

How to prove it?