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

Sum

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

Solution

It is given that ∠BAC = 30° and AB is diameter.

"AB"/2 = 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,

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.

Is there an error in this question or solution?
Chapter 9: Circles - Exercise 9.4 [Page 111]

APPEARS IN

NCERT Exemplar Mathematics Class 10
Chapter 9 Circles
Exercise 9.4 | Q 8 | Page 111
RD Sharma Class 10 Maths
Chapter 8 Circles
Exercise 8.2 | Q 20 | Page 35

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