From an external point \(B\) of a circle with centre \(O\), two tangents \(BC\) and \(BD\) are drawn such that \(\angle DBC=120^\circ\). Prove that \(BC+BD=BO\) (equivalently, \(BO=2\,BC\)).
\(BO=2\,BC\) and since \(BC=BD\), \(BC+BD=BO\).
Step 1: Join \(OC\) and \(OD\), where \(O\) is the centre of the circle and \(C, D\) are the points of tangency.
Step 2: A radius drawn to the point of contact is always perpendicular to the tangent. So, \(OC \perp BC\) and \(OD \perp BD\). That means \(\angle OCB = 90^\circ\) and \(\angle ODB = 90^\circ\).
Step 3: The angle between the tangents at \(B\) is given as \(\angle DBC = 120^\circ\). The angle at the centre, \(\angle COD\), is supplementary to this (they add up to \(180^\circ\)). So, \(\angle COD = 180^\circ - 120^\circ = 60^\circ\).
Step 4: Tangents drawn from an external point are equal in length. Hence, \(BC = BD\). Because of this symmetry, the line \(OB\) will bisect the angle between \(BC\) and \(BD\). So, \(\angle OBC = 60^\circ\).
Step 5: Now consider right triangle \(\triangle OBC\). Its angles are: - \(\angle OCB = 90^\circ\) (because radius is perpendicular to tangent), - \(\angle OBC = 60^\circ\), - So, \(\angle BOC = 30^\circ\).
Step 6: In a right triangle with angles \(30^\circ, 60^\circ, 90^\circ\), the sides are in the ratio: \(1 : \sqrt{3} : 2\). - Side opposite \(30^\circ = 1\), - Side opposite \(60^\circ = \sqrt{3}\), - Hypotenuse = 2.
Step 7: In \(\triangle OBC\): - \(BC\) is opposite \(30^\circ\), - \(OB\) is the hypotenuse. Therefore, \(OB = 2 \times BC\).
Step 8: Since \(BC = BD\), we get: \(BC + BD = BC + BC = 2BC = OB\).
Hence proved: \(BC + BD = BO\).