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.
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.
In a triangle PQR; QR = PR and ∠P = 36o. Which is the largest side of the triangle?
In the following figure ; AC = CD; ∠BAD = 110o and ∠ACB = 74o.
Prove that: BC > CD.
Name the smallest angle in each of these triangles:
In ΔPQR, PQ = 8.3cm, QR = 5.4cm and PR = 7.2cm
Prove that the perimeter of a triangle is greater than the sum of its three medians.
In ΔPQR, PR > PQ and T is a point on PR such that PT = PQ. Prove that QR > TR.
In ΔPQR is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.
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.
