**Given:**

Rhombus is basically a parallelogram with following properties;

(a) All sides are equal.

i.e. AB = BC = CD = DA

(b) Opposite sides are parallel.

(c) Diagonals bisect each other.

i.e. OA = OC and OD = OB

**To Prove:**

Prove that diagonals intersect at right angles.

i.e. ∠AOB = 90 degree

**Proof:**

Consider triangle AOD and COD;

OA = OC {Diagonals of parallelogram bisect each other}

OD = DO {Common Side}

AD = DC {All sides of rhombus are equal}

By SSS congruency, both triangles AOD & COD are congruent.

This gives ∠AOD = ∠COD

Note that both angles ∠AOD & ∠COD form linear pairs.

So we can write;

∠AOD + ∠COD = 180

∠AOD + ∠AOD = 180

2 ∠AOD = 180

∠AOD = 90 degree

Since both ∠AOD & ∠COD are equal, we can write;

∠AOD = ∠COD = 90 degree

Hence, we proved that diagonals of rhombus are perpendicular to each other.