Q. 104.2( 12 Votes )

AB is a diameter of a circle and AC is its chord such that BAC = 30°. If the tangent at C intersect AB extends at D, then BC = BD.

Answer :

True

Given : AB is a diameter of circle with center O and AC is a chord such that BAC = 30° Also tangent at C intersects AB extends at D.


To prove : BC = BD



Proof :


OA = OC [radii of same circle]


OCA = OAC = 30° [Angles opposite to equal sides are equal]


ACB = 90° [angle in a semicircle is a right angle.]


OCA + OCB = 90°


30° + OCB = 90°


OCB = 60° [1]


OC = OB [ radii of same circle]


OBC = OCB = 60° [angles opposite to equal sides are equal]


Now, OBC + CBD = 180° [linear pair]


60 + CBD = 180°


So, CBD = 120° [2]


Also,


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


OCD = 90°


OCB + BCD = 90°


60 + BCD = 90


BCD = 30° [3]


In BCD


CBD + BCD + BDC = 180° [Angle sum property of triangle]


120° + 30° + BDC = 180° [From 2 and 3]


BDC = 30° [4]


From [3] and [4]


BCD = BDC = 30°


BC = BD [Sides opposite to equal angles are equal]


Hence Proved !


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