AB is diameter and AC is a chord of a circle such that ∠BAC=30∘. If the tangent at C intersects AB produced in D, prove that BC = BD [4 MARKS]
Concept: 1 Mark
Application: 3 Marks
Solution:
As OA = OC (radius)
∠OAC=∠OCA=30∘
As OC = OB
∠OCB=∠OBC
In △BAC,
∠OAC+∠OCA+∠OCB+∠OBC=180∘
30∘+30∘+2∠OCB=180∘
∠OCB=60∘
In △BOC,
∠OCB+∠OBC+∠BOC=180∘
60∘+60∘+∠BOC=180∘
∠BOC=60∘
As the tangent at any point of a circle is perpendicular to the radius through the point of contact
∠OCD=90∘
∠BCD=∠OCD−∠BCO
∠BCD=90∘–60∘
∠BCD=30∘ …… (i)
In △OCD,
∠OCD+∠ODC+∠DOC=180∘
90∘+∠ODC+60∘=180∘
∠ODC=30∘ ….. (ii)
From (i) and (ii)
BC = BD