In figure 3.84, O is the centre of the circle. Seg AB, seg AC are tangent segments. Radius of the circle isr and l(AB) = r , Prove that,
is a square.

Given: AB = r = radius of the circle
Here, AB = AC = r {tangents from the same external point are equal}
And OB = OC = r = radius of the circle.
⇒ ∠ OBA = ∠ OCA = 90° Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.
∵ sides of ABOC are equal and opposite angles are 90° each
Hence, ABOC is a square.
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