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 with angle bisector

Let: .

Let: .

Code:

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

Draw perpendicular to

Draw perpendicular to

Since .[1]

In right triangle

In right triangle

Substitute into [1]:.

Therefore: .