Chapter 8.5, Problem 39SPR

To determine

To prove:

  $\frac{AD}{DB}=\frac{AC}{BC}$

Explanation of Solution

Given information:

  

  $\overline{CD}$ bisects $\angle ACB$ .

By construction:

  $\overline{AE}\parallel \overline{CD}$

Proof:

  

The given information is:

  $\overline{CD}$ bisects $\angle ACB:\overline{AE}\parallel \overline{CD}$

By using triangle proportionality theorem:

  $\frac{AD}{DB}=\frac{EC}{BC}\mathrm{...}\left(1\right)$

Then define the angle bisector:

  $\angle 1\cong \angle 2$

Alternate interior angle theorem:

  $\angle 3\cong \angle 1$

Corresponding angle postulate:

  $\angle 2\cong \angle E$

The transitive property:

  $\angle 3\cong \angle E$

Converse of the isosceles triangle theorem:

  $\overline{EC}\cong \overline{AC}$

Definition of the congruence:

  $EC=AC\mathrm{...}\left(2\right)$

Substitute the value in equation (1):

  $\frac{AD}{DB}=\frac{AC}{BC}$

Hence, this condition is proved.