to the risk of sounding redundant.
well, in a triangle, the midsegment is half of its parallel base, namely in this case BD = ½AE.
well, what is the length of BD anyway?
well, we know its coordinates are -1,3 and 2.5,3
[tex] \bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
B(\stackrel{x_2}{2.5}~,~\stackrel{y_2}{3})\qquad
D(\stackrel{x_1}{-1}~,~\stackrel{y_1}{3})\qquad \qquad
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
BD=\sqrt{(-1-2.5)^2+(3-3)^2}\implies BD=\sqrt{(-3.5)^2+0^2}
\\\\\\
BD=\sqrt{3.5^2}\implies BD=3.5\\\\
-------------------------------\\\\
BD=\cfrac{AE}{2}\implies 2BD=AE\implies 2(3.5)=AE\implies 7=AE [/tex]
