Internal Bisector of an Angle of a Triangle

Internal Bisector of an Angle of a Triangle:

Let us construct a triangle ABC with its two sides AB and AC as 5 cm and 7 cm respectively and the included angle BAC = say, 50°. Now let us draw the bisector of ∠BAC meeting BC at D. We now measure BD and CD and find BD/CD. We observe that BD/CD = 5/7 = AB/AC.

In other words, the bisector of an angle of the triangle divides the opposite side in the ratio of the sides including the angle. We can repeat this activity by drawing several such triangles and drawing the bisector of an angle. Every time we shall obtain the same result. Thus, we observe-

Theorem 1: The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

Theorem 2: (Converse of theorem 1). In a triangle ABC, if D is a point on BC such that BD/DC = BA/AC, prove that AD is the bisector of ∠A.

or

In a triangle ABC, if D is a point on BC such that D divides BC in the ratio AB : AC, then AD is the bisector of ∠A.

or

If a line through one vertex of a triangle divides the opposite sides in the ratio of the other two sides, then the line bisects the angle at the vertex.

Theorem 3: The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle.