Q. 44.8( 20 Votes )

If from an external point B of a circle with center O, two tangents BC and BD are drawn such that DBC = 120°, prove that BC + BD = BO i.e., BO = 2 BC.

Answer :


Given : BC and BD are two tangents drawn from a common external point to a circle with center O such that DBC = 120°


To Prove : BC + BD = BO i.e. BO = 2BC


Proof :


In OBC and ODB


OC = OC [ Radii of same circle ]


OB = OB [Common]


BC = BD [Tangents drawn from an external point to a circle are equal ] [1]


OBC OBD [ By Side Side Criterion ]


CBO = DBO [Corresponding parts of congruent triangles are equal ]


Also,


DBC = 120°


CBO + DBO = 120°


CBO + CBO = 120°


CBO = DBO = 60°


OC BC [Tangent at a point on the circle is perpendicular to the radius through point of contact ]


Therefore,


OBC is a right-angled triangle at C


So we have,






BO = 2BC


From [1] we have


BC = BD


BC + BC = BC + BD


2BC = BC + BD


BO = BC + BD


Hence Proved .


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