NCERT Class 9 Solutions: Triangles (Chapter 7) Exercise 7.5 – Part 2
Perpendicular bisector
To find this point, you will construct three perpendicular bisectors, one for each side of the triangle. The point where all three perpendicular bisectors intersect is called the Circumcenter. Using this center point, we can draw a circle that passes through all three vertices.
2. In the triangle located a point in its interior which is equidistant from all the side of the triangle.
Solution:

Draw bisectors of ∠B and ∠ .

Let these angle bisectors intersect each other at point I.

Draw IK BC

Also draw IJ AB and IL AC.

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In and ,
∠IKB ∠IJB = [By construction]
∠IBK ∠IBJ
[ BI is the bisector of ∠B (By construction)]
[Common]
[AngleSideAngle criteria of congruency]
[By Corresponding Parts Congruent Triangles.] ……….equation(1)

Similarly, CIK CIL
[By Corresponding Parts Congruent Triangles.] ……….equation(2)

From eq (1) and (2),

Hence, I is the point of intersection of angle bisectors of any two angles of equidistant from its sides.

Draw bisectors of and .

Let these angle bisectors intersect each other at point I.

Draw

Also draw and .

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In
[By construction]
[ BI is the bisector of B (By construction)]
[Common]
[AngleSideAngle criteria of congruency]
[By Corresponding Parts of Congruent Triangles.] ……….equation(1)

Similarly
[By Corresponding Parts Congruent Triangles.] ……….equation(2)

From eq (1) and (2),

Hence, I is the point of intersection of angle bisectors of any two angles of equidistant from its sides.