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

**Solution:**

The point which is equidistant from all the sides of a triangle is called the incentre of the triangle.

The incentre of a triangle is the intersection point of the angle bisectors of the interior angles of that triangle.

Here, in ∆ABC, we can find the incentre of this triangle by drawing the angle bisectors of the interior angles of this triangle. I is the point where these angle bisectors are intersecting each other.

Therefore, I is the point which is equidistant from all the sides of ∆ABC that is IP = IQ = IP.

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 7

**Video Solution:**

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

NCERT Maths Solutions Class 9 Chapter 7 Exercise 7.5 Question 2

**Summary:**

In ΔABC, a point located in its interior which is equidistant from all the sides of the triangle is the Incenter(I).

**☛ Related Questions:**

- AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50). Show that ∠A > ∠C and ∠B > ∠D.
- In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.
- Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
- ABC is a triangle. Locate a point in the interior of ∆ABC which is equidistant from all the vertices of ∆ABC.

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