# Rhombus and its Theorems

In this section we will discuss rhombus and its theorems.
Rhombus is a parallelogram with all sides equal and parallel.

Rhombus and its Theorems :

Theorem 1 : The diagonals of a rhombus are perpendicular to each other. Given : A rhombus ABCD whose diagonals are AC and BD intersect at O.

Prove that : ∠BOC = ∠DOC = ∠AOD = ∠AOB = 90 0

 Statements Reasons 1) ABCD is a rhombus. 1) Given 2) AB = BC = CD = DA 2) Properties of rhombus. 3) OB = OD and OA = OC 3) As Parallelogram is a rhombus so diagonal bisect each other. 4) BO = OD 4) From (3) 5) BC = DC 5) Properties of rhombus. 6) OC = OC 6) Reflexive (common side) 7) ΔBOC ≅ ΔDOC 7) SSS Postulate. 8) ∠BOC = ≅ ∠DOC 8) CPCTC 9)∠BOC + ∠DOC = 180 9) Linear pair angles are supplementary. 10) 2∠BOC = 180 10) Addition property 11) ∠BOC = 90 11) Division property 12) ∠BOC = ∠DOC = 90 12) As these two angles are congruent.

Hence, ∠AOB = ∠BOC = ∠COD = ∠DOA = 90
0

Converse of the above theorem is also true ⇒ If the diagonals of a parallelogram are perpendicular, then it is a rhombus.

Example :

1) ABCD is a rhombus with ∠ABC = 56. Determine ∠ACD. Solution : ABCD is a rhombus.

∠ ABC = ∠ADC ( Opposite angles are equal)

0

∴ ∠ ODC = ½ ∠ADC ( Diagonals of rhombus bisects the angle)

⇒ ∠ODC = ½ x 56

⇒ ∠ODC = 28
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∠OCD + ∠ODC + ∠COD = 180 ( In ΔOCD, sum of all the angles in a triangle is 180)

∠OCD + 28 + 90 = 180

⇒ ∠OCD + 118 = 180

⇒ ∠OCD = 180 -118

∠OCD = 62
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