# Prove rhombus

by powp
Tags: prove, rhombus
 HW Helper 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 $$\vec a + \vec b$$ 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