check the picture below.
the quadrilateral is a parallelogram, thus the diagonals bisect each other, so, we could have used either diagonal, and get the same midpoint.
[tex]\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
B&({{ 4}}\quad ,&{{ 5}})\quad
% (c,d)
D&({{ -3}}\quad ,&{{ -3}})
\end{array}\qquad
% coordinates of midpoint
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left(\cfrac{{{ -3+4}} }{2}\quad ,\quad \cfrac{-3+5}{2} \right)[/tex]
