In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC

Solution:

Let's represent a diagram according to the given question.

Let AE be the angle bisector of ∠A.

We need to prove that ED is the perpendicular bisector of BC.

∠BAE = ∠CAE ....... (1) [Since, AE is the angle bisector of ∠A]

Now, ∠EBC = ∠CAE ....... (2) [Angles subtended by the same arc EC] 

Also, ∠ECB = ∠BAE ......... (3) [Angles subtended by the same arc BE]

But we know that, ∠BAE = ∠CAE [From equation (1)]

Hence, ∠EBC = ∠ECB [From equations (2) and (3)]

Therefore, BE = EC [Sides opposite to equal angles are equal]

Thus, point E is equidistant from the points B and C. This is only possible when E lies on the perpendicular bisector of BC.

Thus, ED is the perpendicular bisector of BC.

Therefore, the perpendicular bisector of side BC and the angle bisector of ∠A meet on the circumcircle of triangle ABC at point E.

☛ Check: Class 9 Maths NCERT Solutions Chapter 10

Video Solution:

In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC

Maths NCERT Solutions Class 9 Chapter 10 Exercise 10.6 Question 10

Summary:

It is given that in any triangle ABC, the angle bisector of ∠A and perpendicular bisector of BC intersect. We have proved that they intersect on the circumcircle of the triangle ABC.

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