if bisector of any angle of triangle bisects he opposite side then it is isoceles triangle
What is you question?
To determine if a triangle is isosceles when the bisector of any angle bisects the opposite side, follow these steps:
1. Draw the given triangle and label its vertices as A, B, and C.
2. Suppose an angle, ∠B, is bisected by a line segment, which we'll call the angle bisector. Let the angle bisector intersect the side opposite to ∠B at a point D.
3. Assume that the angle bisector divides side AC into two equal segments, AD and DC.
4. Now, to prove that the triangle is isosceles, we need to show that side AB is equal to side BC.
5. Begin the proof by using the Angle Bisector Theorem. According to this theorem, in any triangle, an angle bisector divides the opposite side into segments that are proportional to the other two sides.
6. In triangle ABC, since the angle bisector divides side AC into AD and DC, we can write:
AD/DB = AC/BC (Angle Bisector Theorem)
7. In this case, we are given that AD = DC. Therefore, we can substitute those values into the equation:
AD/DB = AC/BC
DC/DB = AC/BC
1 = AC/BC (Substituting AD = DC)
BC = AC (Cross multiplying)
Since we have proven that side AB is equal to side BC, the triangle is indeed isosceles.
Thus, if the bisector of any angle of a triangle bisects the opposite side, the triangle is isosceles.