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If from an extrenal point B of a circle with centre 0, two tangents BC and BD are drawn such that `angleDBC=120^(@)`, prove that `BC+BD=BO` i.e., BO=2BC.

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To prove that \( BC + BD = BO \) and \( BO = 2BC \) given that \( \angle DBC = 120^\circ \), we can follow these steps: ### Step 1: Draw the Figure Draw a circle with center \( O \) and an external point \( B \). From point \( B \), draw two tangents \( BC \) and \( BD \) touching the circle at points \( C \) and \( D \) respectively. ### Step 2: Identify the Properties of Tangents From the properties of tangents to a circle from an external point, we know that: - \( BC = BD \) (the lengths of tangents from a point outside the circle are equal). ...
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