measures of all angles of the parallelogram.
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Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO = 12
AB 13 then show that ABCD is a rhombus.
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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.
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