**Angle Bisector Theorem :**

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

In ΔABC, AD is the internal bisector

AB/AC = BD/CD

**Converse of Angle Bisector Theorem :**

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

ABC is a triangle. AD divides BC in the ratio of the sides containing the angles <A to meet BC at D.

That is AB/AC = BD/DC

**Example 1 :**

In ΔABC, AD is the bisector of ∠A meeting side BC at D, if AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC

AB/AC = DB/CD

Let DB = x, then CD = 6 - x

10/14 = x/(6-x)

5/7 = x/(6-x)

5(6 - x) = 7x

30 - 5x = 7x

7x + 5x = 30

12x = 30

x = 30/12 = 5/2 = 2.5

BD = 2.5 cm and CD = 6 - 2.5 = 3.5 cm

**Example 2 :**

Check whether AD is bisector of ∠A of ΔABC in each of the following

(i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm.

**Solution :**

By angle bisector theorem :

AB/AC = DB/CD

5/10 = 1.5/3.5

1/2 = 3/7

Hence AD is not the bisector of <A.

(ii) AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm.

**Solution :**

By angle bisector theorem :

AB/AC = DB/CD

4/6 = 1.6/2.4

2/3 = 2/3

Hence AD is the bisector of <A.

**Example 3 :**

In figure <QPR = 90° , PS is its bisector. If ST ⊥ PR, prove that ST × (PQ + PR) = PQ × PR.

**Solution :**

**Example 4 :**

ABCD is a quadrilateral in which AB=AD, the bisector of <BAC and <CAD intersect the sides BC and CD at the points E and F respectively. Prove that EF BD .

**Solution :**

In triangle ABC, AE is the bisector of <BAC

AC/AB = CE/BE ----(1)

In triangle ACD, AF is the bisector <CAD

AC/AD = CF/FD

AC/AB = CF/FD ----(2)

(1) = (2)

CE/BE = CF/FD

Hence EF is parallel to BD.

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