In figure 3.84, 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.

Answer: All four sides of □ABOC equal r and \(\angle ABO = 90^\circ\), so it is a square.

• AB = r (given) and AC = AB = r (tangent segments from an external point are congruent).
• OB = OC = r (radii).

So AB = BO = OC = CA = r, and □ABOC is a rhombus.

\(\angle ABO = 90^\circ\) (tangent is perpendicular to the radius at the point of contact).

A rhombus with one angle \(90^\circ\) is a square. Hence □ABOC is a square.