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.

Answer 1

Proof: Draw segment OB and OC.

`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.