the midsegment that's parallel to AB will be half-way of AC and half-way of BC, Check the picture below.
[tex]\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{2}~,~\stackrel{y_1}{7})\qquad C(\stackrel{x_2}{7}~,~\stackrel{y_2}{6}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{7+2}{2}~~,~~\cfrac{6+7}{2} \right)\implies \left(\cfrac{9}{2}~~,~~\cfrac{13}{2} \right)\implies \blacktriangleright \left( 4\frac{1}{2}~~,~~6\frac{1}{2} \right) \blacktriangleleft \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ B(\stackrel{x_1}{2}~,~\stackrel{y_1}{2})\qquad C(\stackrel{x_2}{7}~,~\stackrel{y_2}{6}) \\\\\\ \left( \cfrac{7+2}{2}~~,~~\cfrac{6+2}{2} \right)\implies \left( \cfrac{9}{2}~~,~~\cfrac{8}{2} \right)\implies \blacktriangleright \left( 4\frac{1}{2}~~,~~4 \right) \blacktriangleleft[/tex]
