Angle bisector theorem and its converse — lesson. Mathematics State Board, Class 10.

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Angle Bisector theorem

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

Given: In

$\triangle ABC$ ,

$AD$ is the internal bisector.

To prove:

$\frac{AB}{AC} = \frac{BD}{CD}$

Construction: Draw a line through

$C$ parallel to

$AB$ . Extend

$AD$ to meet line

$C$ at

$E$ .

Proof: Here,

$CE$ and

$AB$ are two parallel lines cut by a transversal line

$AE$ .

Then,

$\angle AEC = \angle BAE = \angle 1$ [Alternate angles are equal]

Since

$AD$ is the angle bisector, then

$\angle BAD = \angle DAC$

$\triangle ACE$ ,

$\angle CAE = \angle CEA$

Therefore,

$\triangle ACE$ is isosceles.

Thus,

$AC = CE$ ---- (

$1$ )

By AA similarity, we have:

$\triangle ABD \sim \triangle ECD$

$\frac{AB}{CE} = \frac{BD}{CD}$

$\frac{AB}{AC} = \frac{BD}{CD}$ [Using equation (

$1$ )]

Hence, we proved.

The converse of the angle bisector theorem

Statement: 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.

Given:

$ABC$ is a triangle.

$AD$ divides

$BC$ in the ratio of the sides containing the angles

$\angle A$ to meet

$BC$ at

$D$ .

That is,

$\frac{AB}{AC} = \frac{BD}{DC}$ ---- (

$1$ )

To prove:

$AD$ bisects

$\angle A$ . That is,

$\angle 1 = \angle 2$

Construction: Draw

$CE \parallel DA$ . Extend

$BA$ to meet at

$E$ .

Proof: Let us assume

$\angle BAD = \angle 1$ and

$\angle DAC = \angle 2$ .

Since

$DA \parallel CE$ and

$AC$ is the transversal, then, we have:

$\angle BAD = \angle AEC = \angle 1$ [Corresponding angles are equal]

$\angle DAC = \angle ACE = \angle 2$ [Alternate angles are equal]

Consider

$\triangle BCE$ . By Thales theorem, we have:

$\frac{BA}{AE} = \frac{BD}{DC}$ ---- (

$2$ )

From equations (

$1$ ) and (

$2$ ), we have:

$\frac{AB}{AC} = \frac{BA}{AE}$

$AC = AE$ [Cancelling

$AB$ ] ---- (

$3$ )

Therefore,

$\triangle ACE$ is isosceles by equation (

$3$ ).

$\Rightarrow \angle 1 = \angle 2$

Since

$\angle BAD = \angle 1$ and

$\angle DAC = \angle 2$ , then

$AD$ bisects

$\angle A$ .

Hence, we proved.

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3. Angle bisector theorem and its converse

Theory: