show that the diagnols of a rhombus bisects eachother
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Given that ABCD is a rhombus then AB = BC = CD = AD.
To prove that diagonals AC and BD bisect each other.
Proof : Given AB = CD.
In ΔAOB and ΔAOD
AO = AO (Common )
AB = AD (Given )
∠AOB + ∠AOD = 180° ( linear pair )
But ∠AOB = ∠AOD ( Vertically opposite angles)
∠AOB + ∠AOB = 180°
2∠AOB = 180° / 2
∠AOB = 90°
∴The diagonals AC and BD bisect each other
To prove that diagonals AC and BD bisect each other.
Proof : Given AB = CD.
In ΔAOB and ΔAOD
AO = AO (Common )
AB = AD (Given )
∠AOB + ∠AOD = 180° ( linear pair )
But ∠AOB = ∠AOD ( Vertically opposite angles)
∠AOB + ∠AOB = 180°
2∠AOB = 180° / 2
∠AOB = 90°
∴The diagonals AC and BD bisect each other
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