Question

"Question 2 In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Class 9 - Math - Triangles Page 133"

Answer 1

[fig. is in the attachment]

Solution:

Suppose BN & CM are the bisectors of ABC & ACB interest AC & AB at N & M .
Since,O lies on the bisector of BN of ABC, so O will be equidistant from BA and BC.

Again, O lies on the bisector CM of ACB.so O will be equidistant from CA and BC.

Thus, will be equidistant from AB, BC & CA.


Hence, O is a circumcenter of
∆ ABC.

=========================================

Hope this will help you....

Answer 2

Hi friend ...

Solution-- Let BE and CF be the bisectors of angle ABC and angle ACB respectively interacting AC and AB at E and F respectively.

Since O lies on BE ,the bisector of angle ABC ,hence O will be equidistant from AB and BC .Again ,O lies on the bisector CF of angle ACB .Hence, O will be equidistant from BC and AC .Thus, O will be equidistant from AB,BC,and CA.

Hope it will helpful....