Question

In the given figure O is the centre of the circle. AB and AC are tangents drawn from A and BA$\perp$CA. Prove that $\square$BACO is a square

Answer 1

$$\begin{gathered} \angle \mathrm{ACO}=90°(\mathrm{Tangent}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{radius}) \\ \angle \mathrm{ABO}=90°(\mathrm{Tangent}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{radius}) \\ \mathrm{In}\mathrm{quadrialteral}\mathrm{BACO},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{ACO}+\angle \mathrm{ABO}+\angle \mathrm{BAC}+\angle \mathrm{BOC}=360°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow \angle \mathrm{BOC}=360°-\angle \mathrm{ACO}-\angle \mathrm{ABO}-\angle \mathrm{BAC} \\ \Rightarrow \angle \mathrm{BOC}=360°-90°-90-90°=90° \end{gathered}$$

Since all angles of $\square$BACO are 90$°$, it is rectangle.

Then we have:
AB = OC ...(1) (Opposite sides of a rectangle are equal)
AC = OB ...(2) (Opposite sides of a rectangle are equal)
AB = AC ...(3) (Tangents from a external point are equal)
OB = OC ...(4) (Radius of the circle)

From equations (1), (2), (3) and (4), we have:
AB = AC = OC = OB
We know that if all the sides of the rectangle are equal, then it is a square.
∴ $\square$BACO is a square.