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]