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Question
If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that ∠DBC = 120°, prove that BC + BD = BO, i.e., BO = 2BC.
Solution
To prove: BO = 2BC
Given, ∠DBC = 120°
Join OC, OD and BO.
Since, BC and BD are tangents.
∴ OC ⊥ BC and OD ⊥ BD
We know, OB is the angle bisector of ∠DBC.
∴ ∠OBC = ∠DBO = 60°
In right-angled ∆OBC,
cos 60° = `("BC")/("OB")`
⇒ `1/2 = ("BC")/("OB")`
⇒ OB = 2 BC
Also, BC = BD ...[Tangent drawn from external point to circle are equal]
OB = BC + BC
⇒ OB = BC + BD
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