Hello, MATNTRNG!

Here is one form of the proof . . .

Prove that a angle bisector of a triangle separates the opposite side into segments

whose lengths have the same ratio as the ratio of the lengths of the other two sides.

We have $\displaystyle \Delta ABC$ with angle bisector $\displaystyle AD.$

Let: .$\displaystyle \angle BAD = \angle CAD = \theta$

Let: .$\displaystyle AB = c,\;AC = b$

Code:

B
*
* :*
c * : *
* h: *
* : *
* θ : * D F
A * . - - - - - - - + - - o - - - - - +
* θ E * :
* * :
* * :
* * :
b * * :k
* * :
* * :
* * :
* * :
* * :
* *:
o
C

Draw $\displaystyle h = BE$ perpendicular to $\displaystyle AD.$

Draw $\displaystyle k = CF$ perpendicular to $\displaystyle AD.$

Since $\displaystyle \Delta BED \sim \Delta CFD\!:\;\;\dfrac{BD}{DC} \,=\,\dfrac{h}{k}$ .[1]

In right triangle $\displaystyle BEA\!:\;\sin\theta \,=\,\dfrac{h}{c} \quad\Rightarrow\quad h \:=\:c\sin\theta $

In right triangle $\displaystyle CFA\!:\;\sin\theta \,=\,\dfrac{k}{b} \quad\Rightarrow\quad k \,=\,b\sin\theta $

Substitute into [1]:. $\displaystyle \dfrac{BD}{DC} \:=\:\dfrac{c\sin\theta}{b\sin\theta} $

Therefore: .$\displaystyle \dfrac{BD}{DC} \:=\:\dfrac{c}{b}$