AB is a a diameter and AC is a chord of a circle with centre O such that angle BAC =30 the tangent at C intersects extended AB at a point D . prove that BC =BD.

**Given : **A circle with AB as diameter having chord AC.

∠BAC = 30°

Tangent at C meets AB produced at D.

**To prove : **BC = BD

**Construction : ** Join OC

**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°

= 120°

Now, ∠2 + ∠3 = 180° (linear pair)

⇒ 120° + ∠3 = 180°

⇒ ∠3= 60°

AB is diameter of the circle.

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.

We have OC ⊥ CD

⇒ ∠OCD = 90°

⇒ ∠4 + ∠5 (=∠BCD) = 90°

⇒ 60° + ∠5 = 90°

⇒ ∠5 = 30°

In ΔOCD, by angle sum property of Δ

∠5 + ∠OCD + ∠6 = 180°

⇒ 60° + 90° + ∠6 = 18°

⇒ ∠6 + 15° = 180°

⇒ ∠6 = 30

In ΔBCD , ∠5 = ∠6 (= 30°)

⇒ BC = BD (sides opposite to equal angles are equal)

**
**