What I can figure out is that all sides are equal and the oposite corners angles = each other. Not sure what to do
it is the theory which tells you that in a rombus the diagonals are perpendicular so far they bisect each other is it clear?
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] and [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] hence [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)