Question 452127
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I presume what you are trying to do is to prove that the quadrilateral formed by joining all of the adjacent side midpoints of the side of a general rectangle is a rhombus and that the vertices of the original rectangle are the four you specified.


Given my presumption, the four midpoints are


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(0,\,\frac{b}{2}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{a}{2},\,b\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(a,\,\frac{b}{2}\right)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{a}{2},\,0\right)]


To prove a rhombus it is sufficient to prove that all four sides have the same measure.  (This is sufficiently general because a square is a special case of a rhombus).


Use the distance formula to determine the length of one of the sides of the quadrilateral.  Let's begin with the segment defined by the first two endpoints above:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d_1\ =\ sqrt{\left(0\ -\ \frac{a}{2}\right)^2\ +\ \left(\frac{b}{2}\ -\ b\right)^2}\ =\ \frac{\sqrt{a^2\ +\ b^2}}{2}]


Now all you have to do is to show that the distances from the 2nd to the 3rd, 3rd to the 4th, and 4th to the 1st points are all the same as that which we just calculated.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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