Proof of the Angle Bisector Theorem using parallel lines and isosceles triangle properties.

The proof provided in the image is logically sound and correctly demonstrates the Angle Bisector Theorem. Below is a step-by-step verification of the logic to ensure accuracy.

Step 1: Analyze the Construction
* Given: Triangle $ABC$ with angle bisector $AD$. This means $\angle CAD = \angle DAB$.
* Construction: A line is drawn through vertex $B$ parallel to side $AD$, intersecting the extension of side $CA$ at point $E$. So, $AD \parallel BE$.

Step 2: Establish Angle Relationships
* Because $AD \parallel BE$ and line $CE$ acts as a transversal:
* $\angle CAD$ and $\angle AEB$ are corresponding angles. Therefore, $\angle CAD = \angle AEB$.
* Because $AD \parallel BE$ and line $AB$ acts as a transversal:
* $\angle DAB$ and $\angle ABE$ are alternate interior angles. Therefore, $\angle DAB = \angle ABE$.

Step 3: Prove Triangle AEB is Isosceles
* From the given information, we know $\angle CAD = \angle DAB$ (since $AD$ is a bisector).
* From Step 2, we established $\angle CAD = \angle AEB$ and $\angle DAB = \angle ABE$.
* By substitution, $\angle AEB = \angle ABE$.
* Since the base angles of $\triangle AEB$ are equal, the triangle is isosceles. Consequently, the sides opposite these angles are equal in length: $AE = AB$.

Step 4: Apply the Side-Splitting Theorem (Basic Proportionality Theorem)
* In $\triangle CBE$, the line segment $AD$ is parallel to the base $BE$ ($AD \parallel BE$).
* According to the Side-Splitting Theorem (or Thales's Theorem), if a line parallel to one side of a triangle intersects the other two sides, it divides them proportionally.
* Therefore, the ratio of the segments on the left side equals the ratio of the segments on the bottom side:
$$ \frac{AC}{AE} = \frac{DC}{DB} $$
*(Note: The vertices correspond as $C \to C$, $A \to E$, $D \to B$. So side $CA$ corresponds to $CE$? No, the theorem states $\frac{CA}{AE} = \frac{CD}{DB}$ when looking at the segments created by the parallel line cutting the sides $CE$ and $CB$. Specifically, $\frac{Upper Left}{Lower Left} = \frac{Upper Right}{Lower Right}$ relative to the parallel cut. Here, $A$ splits $CE$ and $D$ splits $CB$. So $\frac{CA}{AE} = \frac{CD}{DB}$ is correct.)*

Step 5: Final Substitution
* We have the proportion: $\frac{AC}{AE} = \frac{DC}{DB}$.
* From Step 3, we proved that $AE = AB$.
* Substitute $AB$ for $AE$ in the denominator of the left side:
$$ \frac{AC}{AB} = \frac{DC}{DB} $$

Conclusion:
The steps follow standard geometric principles. The construction creates an isosceles triangle that allows the substitution of side $AB$ for segment $AE$, leading directly to the required proof.

Final Answer:
The proof is correct. By constructing a line parallel to the angle bisector $AD$ through vertex $B$, we create an isosceles triangle $\triangle AEB$ where $AE = AB$. Using the Side-Splitting Theorem on $\triangle CBE$, we get $\frac{AC}{AE} = \frac{DC}{DB}$. Substituting $AB$ for $AE$ yields the final result: $\frac{AC}{AB} = \frac{DC}{DB}$.