Question

AB is a diameter and AC is chord of a circle with centre O such that $\angle BAC={30}^{\circ }$. The tangent at C intersects AB at a point D. Prove that BC =BD.

Answer 1

Proof :

In Δ AOC,

⇒OA = OC [Radii of same circle]

⇒ ∠1 = ∠BAC [Angles opposite to equal sides are equal]

⇒∠1 = 30°

By angle sum property of Δ,

We have,

∠2 = 180° – (30° + 30°)

= 180° – 60°

∠2 = 120°

Now,

∠2 + ∠3 = 180° (linear pair)

⇒ 120° + ∠3 = 180°

⇒ ∠3= 60°

AB is diameter of the circle. [Given]

As,we know that angle in a semi circle is 90°.

⇒ ∠ACB = 90°

⇒ ∠1 + ∠4 = 90°

⇒ 30° + ∠4 = 90°

⇒ ∠4 = 60°

Consider OC is radius and CD is tangent to circle at C.

As, OC ⊥ CD

⇒ ∠OCD = 90°

⇒ ∠4 + ∠5 (=∠BCD) = 90°

⇒ 60° + ∠5 = 90°

⇒ ∠5 = 30°

In ΔOCD,

∠5 + ∠OCD + ∠6 = 180° [By angle sum property of Δ]

⇒ 60° + 90° + ∠6 = 18°

⇒ ∠6 + 15° = 180°

⇒ ∠6 = 30°

Now,

In ΔBCD ,

∠5 = ∠6 [= 30°]

⇒ BC = CD [sides opposite to equal angles are equal]