Advertisements
Advertisements
Question
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Advertisements
Solution
The point which is equidistant from all the sides of a triangle is called the incentre of the triangle. Incentre of a triangle is the intersection point of the angle bisectors of the interior angles of that triangle.

Here, in ΔABC, 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 equidistant from all the sides of ΔABC.
RELATED QUESTIONS
In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.
From the following figure, prove that: AB > CD.

In the following figure, ∠BAC = 60o and ∠ABC = 65o.

Prove that:
(i) CF > AF
(ii) DC > DF
Name the smallest angle in each of these triangles:
In ΔABC, AB = 6.2cm, BC = 5.6cm and AC = 4.2cm
ΔABC is isosceles with AB = AC. If BC is extended to D, then prove that AD > AB.
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PR > PS
In the given figure, T is a point on the side PR of an equilateral triangle PQR. Show that PT < QT
Prove that in an isosceles triangle any of its equal sides is greater than the straight line joining the vertex to any point on the base of the triangle.
In ΔABC, AE is the bisector of ∠BAC. D is a point on AC such that AB = AD. Prove that BE = DE and ∠ABD > ∠C.
