Proving Diagonal Bisection of Rhombus

[powp]

Hello

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

Thanks

 

[A_I_]

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

 

[powp]

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

 

[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?

 

[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)