Proving Diagonal Bisection of Rhombus

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Discussion Overview

The discussion revolves around proving that the diagonals of a rhombus bisect each other. Participants explore various approaches to this geometric property, including theoretical assertions and vector-based proofs.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant asks how to prove that the diagonals of a rhombus bisect each other.
  • Another suggests that proving the diagonals form right angles leads to the conclusion of bisection.
  • A different participant notes that all sides of a rhombus are equal and that opposite angles are equal, expressing uncertainty about the proof process.
  • One participant states that the theory indicates the diagonals of a rhombus are perpendicular and thus bisect each other, seeking clarity on this point.
  • A more detailed approach is presented using vectors, where a participant explains that the diagonals of a parallelogram bisect each other, with the rhombus being a specific case. They provide a mathematical derivation to show that the diagonals bisect each other using vector equations.

Areas of Agreement / Disagreement

Participants express varying degrees of certainty and methods regarding the proof. While some assert theoretical properties, others provide mathematical derivations, indicating that multiple approaches exist without a clear consensus on a singular method.

Contextual Notes

The discussion includes assumptions about the properties of rhombuses and parallelograms, as well as the use of vectors, which may depend on specific definitions and mathematical frameworks. Some steps in the vector proof are not fully resolved.

powp
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Hello

How do you prove that the diagonals od a rhombus bisect each other??

Thanks
 
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when u prove that they make right angles between them.
then, you will get the answer.
180/2 = 90!
 
What I can figure out is that all sides are equal and the oposite corners angles = each other. Not sure what to do
 
it is the theory which tells you that in a rombus the diagonals are perpendicular
so far they bisect each other
is it clear?
 
You can actually prove a stronger result easily using vectors. The diagonals of a parallelogram bisect one another. The rhombus is just a special case of a parallelogram with all sides being equal.

Let the parallelogram be drawn on a Cartesian plane as shown in the diagram, and the sides labelled as vectors as shown. You can see that one diagonal is [tex]\vec a + \vec b[/tex] and the other is [tex]\vec b - \vec a[/tex]

Let [tex]\vec{WO}[/tex] be [tex]k_1(\vec a + \vec b)[/tex]

and

[tex]\vec{OZ}[/tex] be [tex]k_2(\vec b - \vec a)[/tex]

where [tex]k_1, k_2[/tex] are some scalars (which we are to determine).

In triangle WOZ, you can further see that

[tex]\vec{WZ} = \vec{WO} + \vec{OZ}[/tex]

hence

[tex]\vec b = k_1(\vec a + \vec b) + k_2(\vec b - \vec a)[/tex]

[tex]\vec b = (k_1 - k_2)\vec a + (k_1 + k_2)\vec b[/tex]

giving [tex]k_1 - k_2 = 0[/tex] ---eqn (1)

and [tex]k_1 + k_2 = 1[/tex] ---eqn(2)

Solving those simple simultaneous equations yields [tex]k_1 = k_2 = \frac{1}{2}[/tex]

so you know that the diagonals bisect each other. (QED)
 

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