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Given the coordinates of rhombus wxyz are w 0, 4b, x 2a, 0, y 0, -4, and z -2, 0. Prove the segments joining the midpoints of a rhombus from rectangle. as part of proof, find midpoint of xy

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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
K( (0+2)/(2), (4+0)/(2) )=(1,2) \\ L( (0+2)/(2), (-4+0)/(2) )=(1,-2) \\ M( (0-2)/(2), (-4+0)/(2) )=(-1,-2) \\ N( (0-2)/(2), (4+0)/(2) )=(-1,2).

Now find the vectors
\overrightarrow{KL},
\overrightarrow{LM},
\overrightarrow{MN} and
\overrightarrow{KN}:
\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
that means that
\overrightarrow{KL}\perp \overrightarrow{KN}.

Similarly,
\overrightarrow{LK}\perp \overrightarrow{LM},
\overrightarrow{ML}\perp \overrightarrow{MN},
\overrightarrow{NM}\perp \overrightarrow{NK}.
You prove that KLMN is a recctangle.










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