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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.
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Solution
`l`(AB) = r ...[Given] (i)
AB = AC ...\[\boxed{\text{[Tangent segment theorem]}}\] (ii)
But OB = OC = r ...\[\boxed{\text{[Radii of the same circle]}}\] (iii)
From (i), (ii) and (iii)
AB = \[\boxed{\text{AC}}\] = OB = OC = r
∴ Quadrilateral ABOC is \[\boxed{\text{rhombus}}\]
Similarly, ∠OBA = \[\boxed{90°}\] ...[Tangent Theorem]
If one angle of \[\boxed{\text{rhombus}}\] is right angle, then it is a square.
∴ Quadrilateral ABOC is a square.
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