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Proposition:

Let the diagonals AC and BD of the rhombus ABCD intersect at O. It is required to prove that,
(i) ∠AOB = ∠BOC = ∠COD = ∠DOA = 1 right angle
(ii) AO = CO, BO = DO.


Proof:

Step-1: A rhombus is a parallelogram. Therefore, AO = CO, BO = DO. (Diagonals of a parallelogram bisect each other)

Step-2: Now in ΔAOB and ΔBOC,
AB = BC (sides of a rhombus are equal)
AO = CO
and OB = OB. (common side)
So ΔAOB = ΔBOC.
Therefore, ∠AOB = ∠BOC.
∠AOB + ∠BOC = 1 straight angle = 2 right angles.
∠AOB = ∠BOC = 1 right angle.
Similarly, it can be proved that, ∠COD = ∠DOA = 1 right angle. (Proved)


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