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.`

Explanation

To prove that \( BC + BD = BO \) given that \( \angle DBC = 120^\circ \), we can follow these steps:

### Step 1: Understand the Geometry
We have a circle with center \( O \) and an external point \( B \) from which two tangents \( BC \) and \( BD \) are drawn to the circle. By the properties of tangents, we know that \( BC = BD \).

### Step 2: Identify Angles
Given \( \angle DBC = 120^\circ \), we can denote \( \angle CBA \) (the angle between the two tangents at point \( B \)) as \( 120^\circ \). Therefore, the angle \( \angle ABC \) can be calculated as:
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