LEARNATHON
III

Competition for grade 6 to 10 students! Learn, solve tests and earn prizes!

### Theory:

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 $$BE$$.

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

Since $$AD$$ is the angle bisector, then $$\angle BAD = \angle DAC$$

In $$\triangle ACE$$, $$\angle CAE = \angle CEA$$

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.

Converse of 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.