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    ......[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.