Answer:
[tex]\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right),[/tex] [tex]\left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).[/tex]
Step-by-step explanation:
Let points D, E and F have coordinates [tex](x_D,y_D),\ (x_E,y_E)[/tex] and [tex](x_F,y_F).[/tex]
1. Midpoint M of segment DF has coordinates
[tex]\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right).[/tex]
2. Midpoint N of segment EF has coordinates
[tex]\left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).[/tex]
3. By the triangle midline theorem, midline MN is parallel to the side DE of the triangle DEF, then points M and N are endpoints of the midsegment for DEF that is parallel to DE.
