Question -
Answer -
To prove, BC = BD
Join BC and OC.
Given, ∠BAC = 30°
=> ∠BCD = 30°
[angle between tangent and chord is equal to angle made by chord in the alternate segment]
∠ACD = ∠ACO + ∠OCD
∠ACD = 30° + 90° = 120°
[OC ⊥ CD and OA = OC = radius => ∠OAC = ∠OCA = 30°]
In ∆ACD,
∠CAD + ∠ACD + ∠ADC = 180°
[since, sum of all interior angles of a triangle is 180°]
=> 30° + 120° + ∠ADC = 180°
=> ∠ADC = 180° – (30° + 120°) = 30°
Now, in ∆BCD,
∠BCD = ∠BDC = 30°
=> BC = BD
[since, sides opposite to equal angles are equal]