If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that `angleDBC=120^(@)`, prove that `BC+BD=BO` i.e., BO=2BC.

Explanation

Two tangents BD and BC are drawn from an external point B.

To prove BO=2BC
Given, `angleDBC=120^(@)`
Join OC, OD, and BO.
since, BC and BD are tangents.
`:.OCbotBCand ODbotBD`
We know, OB is a angle bisector of `angleDBC`.
`:.angleOBC=angleDBO=60^(@)` In right angled `DeltaOBC`. `cos60^(@)=(BC)/(OB)`
`rArr(1)/(2)=(BC)/(OB)`
`rArrOB=2BC` Also, BC=BD
[tangent drawn from internal point to circle are equal]
`:.OB=BC+BC`
`rArrOB=BC+BD`