Question

How do you prove that the diagonals of a rhombus are perpendicular?

Answer 1

Proof is below.

Explanation:

Rhombus is a parallelogram with all sides equal to each other.
Therefore, rhombus has all the properties of parallelogram. In particular, diagonals of a parallelogram intersect each other at a point that divides each diagonal in half.

Therefore, assuming we have a rhombus `ABCD` with diagonals `AC` and `BD` intersecting at point `O`, triangles `DeltaABO` and `DeltaCBO` are congruent by three sides:
`AB=CB` as sides of a rhombus;
`BO` is shared by both triangles;
`AO=CO` as two halves of a diagonal `AC`.

From the congruence of these triangles follows that angles `/_AOB` and `/_COB` (lying opposite to equal sides `AB` and `CB`) are equal to each other and, therefore are equal to `90^o`.

As a more detailed description of all the properties of parallelograms and other geometrical objects with all the required proofs of each I can suggest to listen the lectures about this at UNIZOR by following the menu options Geometry - Quadrangles.