If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that ∠DBC = 120°, prove that BC + BD = BO, i.e., BO = 2BC
Solution:
Given, BC and BD are the tangents drawn from an external point B of a circle with centre O.
Also, ∠DBC = 120°
We have to prove that BC + BD = BO, i.e., BO = 2BC.
We know that the radius of the circle is perpendicular to the tangent at the point of contact.
So, OC ⟂ BC and OD ⟂ BD
Considering triangles OCB and ODB,
OB = OB = common side
We know that the tangents drawn through an external point to a circle are equal.
So, BC = BD
Also, ∠OCB = ∠ODB = 90°
By RHS criteria, the triangles OCB and ODB are similar.
By corresponding parts of congruent triangles,
The angles ∠OBC and ∠OBD are equal.
So, ∠OBC = ∠OBD = 60°
Considering triangle OCB,
OCB is a right triangle with C at right angle.
cos 60° = BC/OB
By trigonometric ratio of angles,
cos 60° = 1/2
So, 1/2 = BC/OB
OB = 2BC
Therefore, it is proved that OB = 2BC.
✦ Try This: Two tangents segments PA and PB are drawn to a circle with centre O such that ∠APB = 60°. Prove that OP = 2 AP.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10
NCERT Exemplar Class 10 Maths Exercise 9.3 Problem 3
If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that ∠DBC = 120°, prove that BC + BD = BO, i.e., BO = 2BC
Summary:
If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that ∠DBC = 120°. It is proven that BC + BD = BO, i.e., BO = 2BC
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