Prove that the internal bisector of the angle A of a triangle ABC divides BC in the ratio AB:AC

Prove that the internal bisectors of the angles of a triangle are concurrent

Prove that the angle between internal bisector of one base angle and the external bisector of the other base angle of a triangle is equal to one half of the vertical angle.

In a triangle ABC, D is a point on BC such that AD is the internal bisector of angle A . Let angle B = 2 angle C and CD = AB. Then angle A is

Prove that the angle between internal bisector of one base angle and the external bisector of the other base angle of a triangle is equal to one-half of the vertical angle.

The external angle bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle.

In triangle ABC , the internal bisector of angle A angle B and angle C meet at point I . Prove that : (1)/(2) (AB+ BC + CA) lt AI+ BI + CI

If the bisector of angle A of the triangle ABC makes an angle theta with BC , then sintheta=

In a triangle ABC the sides BC=5, CA=4 and AB=3 . If A(0,0) and the internal bisector of angle A meets BC in D (12/7,12/7) then incenter of triangle ABC is

One angle of a triangle is equal to one angle of another triangle and the bisectors of these two equal angles divide the opposite sides in the same ratio, prove that the triangles are similar.

Let triangle ABC be an isosceles with AB=AC. Suppose that the angle bisector of its angle B meets the side AC at a point D and that BC=BD+AD . Measure of the angle A in degrees, is :