Proving Rhombus Diagonals Intersect at Right-Angles

  • Thread starter masterofthewave124
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In summary, the rhombus has two diagonals that are perpendicular. Fixing the coordinates so that all sides are equal results in a solved problem.
  • #1
masterofthewave124
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i have to prove the diagonals of a rhombus intersect at right-angles using the scalar dot product.

i have set up a cartesian plane system where A lies on the origin. numbering the sequential points clockwise, i let B = (a,b), C = (a+c, b) and D = (c,0). I then thought if i set up vectors AC and DB and found the dot product between them, i could get a value of zero. unfortunately that's not the case so where did i go wrong?
 
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  • #2
Perhaps you made an arithmetic error? Your method appears good.
 
  • #3
Well, he didn't set up a rhombus; he set up a parallelogram.
 
  • #4
Hurkyl said:
Well, he didn't set up a rhombus; he set up a parallelogram.
So that would explain why my dot product was non zero. I fixed it for a rhombus and the proof works fine.
 
  • #5
hmm...where would my flaw lie? i think i just may be overlooking a critical property of a rhombus.
 
  • #6
What's the definition of a rhombus?
 
  • #7
a parallelogram having equal sides. i still can't see how the way i defined the coordinates is wrong, maybe just a little obscure (can be simplified).
 
  • #8
So did you, anywhere in your set-up, impose the condition that all the sides were equal?
 
  • #9
ahh I am confused, the way i have my diagram, a traditional rhombus, it looks right. i can see how a square would be different.

so back to basics:
AB = BC = CD = DA

but that doesn't necessarily mean the coordinates will be represented that way, just the magnitude.
 
  • #10
ahh I am confused, the way i have my diagram, a traditional rhombus, it looks right.
What if you set a = 400, b = 400, and c = 1?


In the way you set up your problem, the vector AB = (a, b), and the vector AD = (c, 0). But clearly they are not automatically equal!
 
  • #11
hmm i see your point. but then what would AB = ?

basically a^2 + b^2 = c^2

so AB would equal (sqrt(c^2-b^2), sqrt (c^2 - a^2)), can't be right, much too messy...
 
  • #12
basically a^2 + b^2 = c^2
I suspect that this is all you need -- go back and redo your work, but make use of this relation.


Incidentally, this problem is easier if you don't use coordinates at all!
 
  • #13
oh ok, i just left my coordinates as is but when i got down to the dot product,

after i get a^2 + ac - ac - c^2 + b^2
i stated that a^2 + b^2 = c^2
so c^2 - c^2 = 0i hope that's sufficient. what other method would have been easier Hurkyl?
 
  • #14
Do essentially the exact same thing, but never use coordinates: leave everything in terms of vectors.
 
  • #15
All right, I'll spoil the answer if you haven't worked it out yet yourself!


Let x and y be the vectors denoting two sides of our rhombus.

Then, as you noted, the two diagonals are:

x + y

and

x - y

and because the sides are equal:

x² = y²

The dot product of the diagonals is:

(x + y).(x - y) = x² - y² = 0

and thus, they are perpendicular.


(Where I've used the notation that x² means x.x)
 

1. How do you prove that the diagonals of a rhombus intersect at right angles?

To prove that the diagonals of a rhombus intersect at right angles, you can use the following theorem: In a rhombus, the diagonals are perpendicular bisectors of each other.

2. Why is it important to prove that the diagonals of a rhombus intersect at right angles?

Proving that the diagonals of a rhombus intersect at right angles is important because it is a fundamental property of a rhombus. It helps to define and differentiate a rhombus from other quadrilaterals.

3. What is the difference between a rhombus and a square?

A square is a special type of rhombus where all sides are equal in length. However, in a rhombus, the angles can vary and are not necessarily all 90 degrees like in a square.

4. Can you prove that the diagonals of a rhombus intersect at right angles without using the theorem?

Yes, another way to prove that the diagonals of a rhombus intersect at right angles is by using the properties of a rhombus. Since a rhombus has four congruent sides, its opposite angles are also congruent. Therefore, the diagonals bisect each other at a 90-degree angle.

5. Is it possible for the diagonals of a rhombus to not intersect at right angles?

No, it is not possible for the diagonals of a rhombus to not intersect at right angles. As mentioned earlier, it is a defining property of a rhombus. If the diagonals do not intersect at right angles, then the figure is not a rhombus.

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