AB is a diameter and AC is a chord of a circle with centre o such that BAC = 30°. The tangent at C intersects AB at a point D. Prove that BC = BD.
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Answer:
AB is a diameter and AC is a chord of a circle. BC = BD is proved
Solution:
The diagram for this sum is attached below
We have 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
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