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.
3 Ресурс 1.png
 
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.
4 Ресурс 1.png
 
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.