Question

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

Answer 1

Answer:

In the fig

given that - AB is diameter and AC is chord of circle such that BAC=30

to prove that BC = BD

∠ACB = 90° (Since AB is the diameter of the circle)

In ∆ACB

∠A + ∠C + ∠B = 180°

30°+ 90° + ∠B = 180°

120° + ∠B = 180°

∠B = 180°- 120°

∠B = 60°

∠CBA = 60°…………….(Equation 1)

Now in ∆OCB,

OC = OB

So ∠OCB = ∠OBC

∴∠OCB = 60°

Now ∠OCD = 90° (Perpendicular from center of tangent)

∠OCB + ∠BCD = 90°

60°+ ∠BCD = 90°

∠BCD = 90°- 60°

∠BCD = 30°………………..(Equation 2)

∠CBO = ∠BCD + ∠CDB

60°=30° + ∠CDB

∠CDB= 30°……………..(Equation 3)

From equation 2 and equation 3

BC =BD

Thus BC = BD is proved