Let us look at the following graph:
Let the mid-point be \(M\) with the vertex \(M(x\), \(y)\).
From the similarity property, it is understood that \(\triangle{AMM'}\) and \(\triangle{MDB}\) are similar.
This makes the two corresponding sides equal.
Therefore, the ratio between the two corresponding sides are also equal.
\(\frac{AM'}{MD} = \frac{MM'}{BD} = \frac{AM}{MB} = \frac{1}{1}\)
We know that since \(M\) is the mid-point of \(AB\), \(AM = MB\).
So, \(\frac{x - x_1}{x_2 - x} = \frac{y - y_1}{y_2 - y} = \frac{1}{1}\)
Let us only consider \(\frac{x - x_1}{x_2 - x} = \frac{1}{1}\)
\(x - x_1 = x_2 - x\)
\(2x = x_1 + x_2\)
\(x = \frac{x_1 + x_2}{2}\)
Similarly \(y\) becomes \(\frac{y_1 + y_2}{2}\)
\(y = \frac{y_1 + y_2}{2}\)
Hence, the mid-point be \(M\) with the vertex \(M(x\), \(y)\) becomes \(M(\frac{x_1 + x_2}{2}\), \(\frac{y_1 + y_2}{2})\).