Prove rhombus

1. Apr 2, 2005

powp

Hello

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

Thanks

2. Apr 2, 2005

A_I_

when u prove that they make right angles between them.
then, you will get the answer.
180/2 = 90!

3. Apr 2, 2005

powp

What I can figure out is that all sides are equal and the oposite corners angles = each other. Not sure what to do

4. Apr 2, 2005

A_I_

it is the theory which tells you that in a rombus the diagonals are perpendicular
so far they bisect each other
is it clear?

5. Apr 2, 2005

Curious3141

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 $$\vec a + \vec b$$ and the other is $$\vec b - \vec a$$

Let $$\vec{WO}$$ be $$k_1(\vec a + \vec b)$$

and

$$\vec{OZ}$$ be $$k_2(\vec b - \vec a)$$

where $$k_1, k_2$$ are some scalars (which we are to determine).

In triangle WOZ, you can further see that

$$\vec{WZ} = \vec{WO} + \vec{OZ}$$

hence

$$\vec b = k_1(\vec a + \vec b) + k_2(\vec b - \vec a)$$

$$\vec b = (k_1 - k_2)\vec a + (k_1 + k_2)\vec b$$

giving $$k_1 - k_2 = 0$$ ---eqn (1)

and $$k_1 + k_2 = 1$$ ---eqn(2)

Solving those simple simultaneous equations yields $$k_1 = k_2 = \frac{1}{2}$$

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

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Last edited: Apr 2, 2005