Question

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

Answer 1

Given :AB is the diameter and AC is the chord of a cirlcle such that angle BAC-30. if the tangent at C intersect AB produced at D
To prove : BC=BD
Construction : Join OC
Proof : In triangle AOC,
OA = OC (radii of same circle)
=> angle 1 = angle BAC (angles opposite to equal sides are equal)
=> angle 1 = 30 degrees
By angle sum property of triangle,
We have, angle 2 = 180 (30 + 30)
= 180 60
= 120
Now, angle 2 + angle 3 = 180 (linear pair)
=> 120 + angle 3 = 180
=> angle 3= 60
AB is diameter of the circle.
We know that angle in a semi circle is 90 degrees .
=> angle ACB = 90
=> angle 1 + angle 4 = 90
=> 30 + angle 4 = 90
=> angle 4 = 60
Consider OC is radius and CD is tangent to circle at C.
We have OC is perpendicular to CD
=> angle OCD = 90
=> angle 4 + angle 5 (=angle BCD) = 90
=> 60 + ?5 = 90
=> angle 5 = 30
In triangle OCD, by angle sum property of triangle
=> angle 5 + angle OCD + angle 6 = 180
=> 60 + 90 + angle 6 = 180
=> angle 6 + 15 = 180
=> angle 6 = 30
In triangle BCD , angle 5 = angle 6 (= 30)
=> BC = CD (sides opposite to equal angles are equal)
Hence Proved