UPSKILL MATH PLUS

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**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.

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.