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Triangle Angle Bisector Theorem

An angle bisector of an angle of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.

By the Angle Bisector Theorem,

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

Proof:

Draw $\stackrel{\leftrightarrow }{BE}\parallel \stackrel{\leftrightarrow }{AD}$ .

Extend $\overline{CA}$ to meet $\stackrel{\leftrightarrow }{BE}$ at point $E$ .

By the Side-Splitter Theorem,

$\frac{CD}{DB}=\frac{CA}{AE}$ ---------( $1$ )

The angles $\angle 4\text{and}\angle 1$ are corresponding angles.

So, $\angle 4\cong \angle 1$ .

Since $\overline{AD}$ is a angle bisector of the angle $\angle CAB,\angle 1\cong \angle 2$ .

By the Alternate Interior Angle Theorem , $\angle 2\cong \angle 3$ .

Therefore, by transitive property, $\angle 4\cong \angle 3$ .

Since the angles $\angle 3\text{and}\angle 4$ are congruent , the triangle $\Delta ABE$ is an isosceles triangle with $AE=AB$ .

Replacing $\mathrm{AE}$ by $\mathrm{AB}$ in equation ( $1$ ),

$\frac{CD}{DB}=\frac{CA}{AB}$