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Question

AB is a diameter and AC is a chord of a circle with centre O such that BAC= 30° . The tangent at C intersects AB at a point D . Prove that BC = BD.

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Solution

It is given that ∠BAC = 30° and AB is diameter.
AB2=OA=OC Radius

ACB = 90° (Angle formed by the diameter is 90°)

In ∆ABC,

ACB + ∠BAC + ∠ABC = 180°

90° + 30° + ∠ABC = 180°

⇒ ∠ABC = 60°

⇒ ∠CBD = 180° – 60° = 120° ( ∠CBD and ∠ABC form a linear pair)

In ∆OCD,

OCD = 90° (Angle made by Radius on the tangent)

OBC = ∠ABC = 60°

Since OB = OC, ∠OCB = ∠OBC = 60° (OC = OB = radius)

In ∆OCB,

⇒ ∠COB + ∠OCB + ∠OBC = 180°

⇒ ∠COB + 60° + 60° = 180°

⇒ ∠COB = 60°

In ∆OCD,

COD + ∠OCD + ∠ODC = 180°

60° + 90° + ∠ODC = 90° (∠COD = ∠COB)

⇒ ∠ODC = 30°

In ∆CBD,

CBD = 120°

BDC = ∠ODC = 30°

⇒ ∠BCD + ∠BDC + ∠CBD = 180°

⇒ ∠BCD + 30° + 120° = 180°

⇒ ∠BCD + 30° = ∠BDC

Angles made by BC and BD on CD are equal, so ∆CBD is an isosceles triangle and therefore, BC = BD.



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