O is the circumcentre of the triangle ABC and D is the mid-point of the base BC. Prove that ∠BOD = ∠A.
Solution:
Given, ABC is a triangle
O is the circumcentre of the triangle
D is the midpoint of the base BC
We have to prove that ∠BOD = ∠A.
Join OB, OC and OD
Considering triangles BOD and COD,
OB = OC = Radius of the circle
Since D is the midpoint of BC
BD = DC
Common side = OD
The Side-Side-Side congruence rule states that, if all the three sides of a triangle are equal to the three sides of another triangle then the triangles are congruent.
By SSS criterion, the triangles BOD and COD are similar.
The Corresponding Parts of Congruent Triangles are Congruent (CPCTC) theorem states that when two triangles are similar, then their corresponding sides and angles are also congruent or equal in measurements.
By CPCTC,
∠BOD = ∠COD
So, ∠BOC = 2 ∠BOD ---------------------- (1)
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.
∠BOC = 2 ∠BAC
From (1),
2∠BOD = 2∠BAC
Therefore, ∠BOD = ∠BAC
✦ Try This: In the adjacent figure, if ∠AOC=110°, then the value of ∠D and∠B respectively.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.3 Problem 7
O is the circumcentre of the triangle ABC and D is the mid-point of the base BC. Prove that ∠BOD = ∠A.
Summary:
O is the circumcentre of the triangle ABC and D is the mid-point of the base BC. It is proven that ∠BOD = ∠A
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