Answer:
Rhombus.
Step-by-step explanation:
Consider quadrilateral ABDC with vertices at points (2,2), (2,-2), (-1,2), (-1,-2). This quadrilateral is a rectangle. The sides' midpoints are:
- E(2,0);
- F(0.5,-2);
- G(-1,0);
- H(0.5,2).
Quadrilateral EFGH is always a parallelogram, because midlines EF and GH are parallel to the diagonal AD (by the triangle's midline theorem) and HE, GF are parallel to the diagonal BC. Thus, EF || GH and HE || GF.
Note that
[tex]EF=FG=GH=HE=\sqrt{(2+0.5)^2+(0+2)^2}=\sqrt{6.25+4}=\sqrt{10.25}\ un.[/tex]
Thus, this parallelogram is rhombus.
