WXYZ is a rhombus with vertices W(0,4b), X(2a,0), Y(0,-4), Z(-2,0). You can see thap points W and Y lie on the y-axis and points X and Z lie on the x-axis. Then the centre of the rhombus is origin, thus W and Y are symmetric about the origin. Then b=1 and point W has coordinates (0,4). Similarly points X and Z are symmetric about the origin and a=1, hence X(2,0).
Let KLMN be middlepoints of segments WX, XY, YZ, ZW, respectively. Then [tex]K( \frac{0+2}{2}, \frac{4+0}{2} )=(1,2) \\ L( \frac{0+2}{2}, \frac{-4+0}{2} )=(1,-2) \\ M( \frac{0-2}{2}, \frac{-4+0}{2} )=(-1,-2) \\ N( \frac{0-2}{2}, \frac{4+0}{2} )=(-1,2)[/tex].
Now find the vectors [tex]\overrightarrow{KL}[/tex], [tex]\overrightarrow{LM}[/tex], [tex]\overrightarrow{MN}[/tex] and [tex]\overrightarrow{KN}[/tex]:[tex]\overrightarrow{KL}=(1-1,-2-2)=(0,-4) \\ \overrightarrow{KN}=(-1-1,2-2)=(-2,0) \\\overrightarrow{ML}=(-1-1,-2-(-2))=(-2,0) \\ \overrightarrow{MN}=(-1-(-1),2-(-2))=(0,4)\\\overrightarrow{KL}\cdot \overrightarrow{KN}=0\cdot (-2)+(-4)\cdot 0=0[/tex]
that means that [tex]\overrightarrow{KL}\perp \overrightarrow{KN}[/tex].
Similarly, [tex]\overrightarrow{LK}\perp \overrightarrow{LM}[/tex],[tex]\overrightarrow{ML}\perp \overrightarrow{MN}[/tex], [tex]\overrightarrow{NM}\perp \overrightarrow{NK}[/tex].
You prove that KLMN is a recctangle.
