Question

AB is a diameter of a circle with centre O and AC is its chord such that ∟BAC=30®. If the tangent drawn at C intersects extend AB at D, then Show that BC=BD

Answer 1

I think you have fig mos probably that's why I've not provided..

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°

 ∠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]