1. We have that B (-2,4) and D (2,1), then the midpoint has coordinates
[tex]\begin{gathered} x=\frac{-2+2}{2}=0 \\ y=\frac{4+1}{2}=\frac{5}{2} \end{gathered}[/tex]and the midpoint is (0,5/2).
On the other hand A (-4,1) and C (4,4), so the midpoint has coordinates
[tex]\begin{gathered} x=\frac{-4+4}{2}=0 \\ y=\frac{1+4}{2}=\frac{5}{2} \end{gathered}[/tex]and the midpoint is (0, 5/2).
In conclusion, the midpoint of segments BD and AC is (0,5/2).
2. To classify the triangle we need to know the length of its sides
[tex]\begin{gathered} \bar{AB}=\sqrt[]{(3-(-2))^2+\mleft(0+2\mright)^2}=\sqrt[]{25+4}=\sqrt[]{29} \\ \bar{BC}=\sqrt[]{(2-(-2))^2+(4-2)^2}=\sqrt[]{16+4}=\sqrt[]{20} \\ \bar{AC}=\sqrt[]{(3-2)^2+(4-0)^2}=\sqrt[]{1+16}=\sqrt[]{17} \end{gathered}[/tex]Since neither of its sides has equal length, then it is not equilateral os isosceles. Besides,
[tex]17^2+20^2\ne29^2[/tex]Then it is not a right triangle.
In conclusion the answer is none of the options