# Prove rhombus

by powp
Tags: prove, rhombus
 P: 93 Hello How do you prove that the diagonals od a rhombus bisect each other?? Thanks
 P: 137 when u prove that they make right angles between them. then, you will get the answer. 180/2 = 90!
 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
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?
 HW Helper P: 2,634 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) Attached Thumbnails

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