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.