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,

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)

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