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Question
In the given figure, O is the centre of the circle. Seg AB, seg AC are tangent segments. Radius of the circle is r and l(AB) = r, Prove that ▢ABOC is a square. 
Proof: Draw segment OB and OC.
l(AB) = r ......[Given] (I)
AB = AC ......[`square`] (II)
But OB = OC = r ......[`square`] (III)
From (i), (ii) and (iii)
AB = `square` = OB = OC = r
∴ Quadrilateral ABOC is `square`
Similarly, ∠OBA = `square` ......[Tangent Theorem]
If one angle of `square` is right angle, then it is a square.
∴ Quadrilateral ABOC is a square.
Solution
Proof: Draw segment OB and OC.
l(AB) = r ......[Given] (I)
AB = AC ......[Tangent segment theorem] (II)
But OB = OC = r ......[Radii of the same circle] (III)
From (I), (II) and (III)
AB = AC = OB = OC = r
∴ Quadrilateral ABOC is rhombus
Similarly, ∠OBA = 90° ......[Tangent Theorem]
If one angle of rhombus is right angle, then it is a square.
∴ Quadrilateral ABOC is a square.
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