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How can we use vectors and dot products to show that the diagonals of a parallelogram intersect at $90^\circ$ if and only if the figure is a rhombus?

I did the proof, but I realized my final answer would be a rectangle. (I know a rhombus is a type of rectangle, too). But I only want to prove the two diagonals are orthogonal.

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  • $\begingroup$ A rhombus is not "a type of rectangle" (unless it's specifically a square). However, both rhombi and rectangles are types of parallelograms. $\endgroup$ – Blue Apr 8 '13 at 22:08
  • $\begingroup$ maybe you compared adjacent sides instead of the diagonals, that's why you came up with a rectangle instead $\endgroup$ – suissidle Apr 8 '13 at 22:12
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Hint:

The diagonals bisect at $90^\circ \implies m_1m_2=-1$, where $m_1$ and $m_2$ are slopes of the diagonals. That's a parallelogram, opposite sides are equal$ \implies$ magnitude of opposite vectors are equal. Now take into one more property of parallelogram, the diagonals bisect, now you can just use Pythagoras to show sides are equal.

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Suppose you have a parallelogram with vertices A,B,C and D and is oriented as such: AB is parallel to CD (hence AD is parallel to BC). To satisfy that this parallelogram is a rhombus, all sides must be equal. Construct a vector that goes from A to C, $\vec{x_{AC}}$ and another that goes from B to D, $\vec{x_{BD}}$. This shape is a rhombus if and only if $\vec{x_{AC}} \cdot \vec{x_{BD}} = 0$ and the length of the two diagonal vectors are not equal to each other. Another thing you can try to show is that the two diagonals are bisectors of one another, but that is another problem.

(Consider the case where the length of the diagonals are equal, what shape would that be?)

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  • $\begingroup$ why are you excluding the possibility of a square? it's a subtype of rhombus (and leavejava's statement clearly includes the square) $\endgroup$ – suissidle Apr 8 '13 at 22:03
  • $\begingroup$ i just prove that shape it's a rectangle.But the statement only want me to prove a rhombus. $\endgroup$ – JavaLeave Apr 8 '13 at 22:07
  • $\begingroup$ yes, my beef was with kvmu for singling out the square as 'not a rhombus', which is not in the intention of the problem $\endgroup$ – suissidle Apr 8 '13 at 22:10
  • $\begingroup$ my answer is like $4*AB\centerdot BC = 0$. which is show $AB\bot CD$ $\endgroup$ – JavaLeave Apr 8 '13 at 22:22
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    $\begingroup$ but maybe it IS the same, who knows? why does it bother you so much? a square is a rhombus just accept it =D $\endgroup$ – suissidle Apr 8 '13 at 23:43
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write each diagonal (vector) in terms of two adjacent sides (vectors) and show their inner product equals $0$ if and only if the two sides have equal length

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  • $\begingroup$ You can assume the four vertices have the form $(0,0)$, $(a, 0)$, $(b,c)$, $(a+b,c)$ with $a, b, c > 0$ and then follow the advice given above. $\endgroup$ – Stefan Smith Apr 8 '13 at 22:13
  • $\begingroup$ More precisely, "... show their inner product equals $0$ if and only if the two sides have equal length." (BTW, you want them to be two adjacent sides.) A parallelogram with congruent adjacent sides is a rhombus. $\endgroup$ – Blue Apr 8 '13 at 22:14
  • $\begingroup$ @StefanSmith: It may be easier to simply use generic vectors $u$ and $v$. The formula for the dot product of the diagonal vectors reduces nicely. $\endgroup$ – Blue Apr 8 '13 at 22:16
  • $\begingroup$ yes, i neglected the other direction of the proof..corrected $\endgroup$ – suissidle Apr 8 '13 at 23:42
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Assume the centroid of the parallelogram $P\subset {\mathbb R}^2$ at the origin. Then the four vertices are $a$, $b$, $-a$, $-b$, and the two sidelengths are $|a-b|$ and $|a+b|$. Since $$|a+b|^2-|a-b|^2=4\ a\cdot b$$ it follows that $P$ is a rhombus iff $a\perp b$.

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