Chapter 7.6, Problem 19WE

To determine

To prove: The triangle angle -bisector theorem.

Explanation of Solution

If Point L and M lie on AB and CD , respectively. If $\frac{AL}{LB}=\frac{CM}{MD}$ , then $\overline{AB}$ and $\overline{CD}$ are divided proportionally.

The two sides are separated proportional if one side is crossed by a parallel to one side.

  $\begin{array}{cc}& \text{If Δ }DEF;\overline{DG}\text{ bisects }\angle FDE\\ & \text{Then, }\frac{GF}{GE}=\frac{DF}{DE}\end{array}$

If $\text{Δ}FKE:\overline{DG}\parallel \overline{EK}$

Then, $\frac{GF}{GE}=\frac{DF}{DK}$

Or,

  $\frac{GF}{GE}=\frac{DF}{DE}$

Hence, if a line splits the angle of a triangle, the opposite side is divided into segments equal to the other two sides.

  $\begin{array}{cc}& \text{If Δ }DEF;\overline{DG}\text{ bisects }\angle FDE\\ & \text{Then, }\frac{GF}{GE}=\frac{DF}{DE}\end{array}$