AB is a a diameter and AC is a chord of a circle with centre O such that angle BAC - Maths - Circles

Question

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

Aakash Sharma · Accepted Answer

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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)

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.