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Prove rhombus

by powp
Tags: prove, rhombus
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Apr2-05, 06:19 AM
P: 93

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

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Apr2-05, 07:21 AM
P: 137
when u prove that they make right angles between them.
then, you will get the answer.
180/2 = 90!
Apr2-05, 07:50 AM
P: 93
What I can figure out is that all sides are equal and the oposite corners angles = each other. Not sure what to do

Apr2-05, 08:02 AM
P: 137
Prove rhombus

it is the theory which tells you that in a rombus the diagonals are perpendicular
so far they bisect each other
is it clear?
Apr2-05, 08:49 AM
HW Helper
Curious3141's Avatar
P: 2,952
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]


[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]


[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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