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Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO show that ABCD is a rhombus.

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Given: AO = 5, BO = 12 and AB = 13. 

To prove: ABCD is a rhombus.

Proof:

AO = 5, BO = 12, AB = 13 [Given] 

AO2 + BO2 = 52 + 122 = 25 + 144 

∴ AO2 + BO2 = 169 …..(i) 

AB2 = 132 = 169 ….(ii) 

∴ AB2 = AO2 + BO2 [From (i) and (ii)] 

∴ ∆AOB is a right-angled triangle. [Converse of Pythagoras theorem] 

∴ ∠AOB = 90° 

∴ seg AC ⊥ seg BD …..(iii) [A-O-C] 

∴ In parallelogram ABCD, 

∴ seg AC ⊥ seg BD [From (iii)] 

∴ ABCD is a rhombus. [A parallelogram is a rhombus perpendicular to each other]

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