The circumcentre of the triangle ABC is O. Prove that ∠OBC + ∠BAC = 90º.

Solution:

Given, ABC is a triangle

O is the circumcentre of the triangle

We have to prove that ∠OBC + ∠BAC = 90º.

Join BO and CO.

Let ∠OBC = ∠OCB = θ --------------------------- (1)

Considering triangle OBC,

By angle sum property,

∠BOC + ∠OCB + ∠CBO = 180º

∠BOC + θ + θ = 180º

∠BOC = 180 - 2θº ---------------------------------- (2)

We know that in a circle the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

So, ∠BOC = 2 ∠BAC

∠BAC = 1/2 ∠BOC

From (2),

∠BAC = 180º - 2θ/2

∠BAC = 90º - θ

∠BAC + θ = 90º

From (1),

∠BAC + ∠OBC = 90º

Therefore, ∠OBC + ∠BAC = 90º

✦ Try This: The distance of the circumcentre of the acute angled ∠ABC from the sides BC,CA and AB are in the ratio ……..

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10

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NCERT Exemplar Class 9 Maths Exercise 10.3 Problem 13

The circumcentre of the triangle ABC is O. Prove that ∠OBC + ∠BAC = 90º.

Summary:

The circumcenter is the center point of the circumcircle drawn around a polygon. The circumcentre of the triangle ABC is O. It is proven that ∠OBC + ∠BAC = 90º

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☛ Related Questions:

• A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in maj . . . .
• In Fig.10.13, ∠ADC = 130° and chord BC = chord BE. Find ∠CBE
• In Fig.10.14, ∠ACB = 40º. Find ∠OAB