  Question

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

Solution

## Given :AB is the diameter and AC is the chord of a circle 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 the 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)  Suggest corrections   