Question

Two tangents segments BC and BD are drawn to a circle with center O such that ∠CBD = 120°. Prove that OB = 2BC

Answer 1

Here, OB is the bisector of ∠CBD.

(Two tangents are equally inclined to the line segment joining the center to that point)

\(\therefore \angle CBO=\angle DBO=\frac{1}{2}\angle CBD=60^\circ\)

∴ From △BOD, ∠BOD = 30°

Now, from right – angled △BOD,

\(\Rightarrow\) \(\frac{BD}{OB}=sin30^\circ\)

\(\Rightarrow\) OB = 2BD

\(\Rightarrow\) OB = 2BC (Since tangents from an external point are equal. i.e., BC = BD)

∴ OB = 2BC

Answer 2

It can be clearly show that OB bisects ∠DBC.

∴∠OBC = ∠OBD = 60

In ?OBC,

∠OBC = 60, ∠OCB = 90

∠COB + ∠OBC +∠OCB = 180 [Angle sum property of triangle]

∠COB + 60 + 90 = 180

∠COB = 180 – 150 = 30

sin(∠COB) = BC/BO

1/2 = BC/BO

Hence, BO = 2BC