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Question
Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO = 12 and AB = 13 then show that `square`ABCD is a rhombus.
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Solution
Given: AO = 5, BO = 12 and AB = 13
To Prove: `square`ABCD is a rhombus.
Proof:
AO2 + BO2
= 52 + 122
= 25 + 144
= 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)]
∴ `square`ABCD is a rhombus. ...[A parallelogram is a rhombus perpendicular to each other]
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