In $△ABC,$ consider the angle bisector $\ell$ that divides $\angle ABC$ into two congruent angles. Let $\angle 1$ and $\angle 2$ be these congruent angles.

By the Parallel Postulate, a parallel line to $\ell$ can be drawn through $A.$ Additionally, if $\overline{BC}$ is extended, it will intersect this line. Let $E$ be their point of intersection.

Let $\angle 3$ be the alternate interior angle to $\angle 1$ formed at $A.$ Also, let $\angle 4$ be the corresponding angle to $\angle 2$ formed at $E.$

By the Corresponding Angles Theorem, $\angle 2$ is congruent to $\angle 4.$ Remember that it is also known that $\angle 1$ is congruent to $\angle 2.$ By the Transitive Property of Congruence, $\angle 1$ and $\angle 4$ are congruent angles. Additionally, by the Alternate Interior Angles Theorem, $\angle 1$ is congruent to $\angle 3.$ Using the Transitive Property of Congruence one more time, it can be said that $\angle 3$ and $\angle 4$ are also congruent angles. This can be shown in the diagram. Note that $△ACE$ is divided by $\ell ,$ which is parallel to $\overline{AE}.$ Therefore, by the Triangle Proportionality Theorem, $\ell$ divides the other two sides of this triangle proportionally. The Converse Isosceles Triangle Theorem states that if two angles in a triangle are congruent, the sides opposite them are congruent. This means that $\overline{EB}$ is congruent to $\overline{AB}.$ Therefore, by the definition of congruent segments, they have the same length. $AB$ can be substituted for $EB$ in the above proportion.