CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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)

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
QuestionImage
QuestionImage
View More...


People also searched for
QuestionImage
QuestionImage
View More...



footer-image