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

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

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

By the Corresponding Angles Theorem, ∠2 is congruent to ∠4. Remember that it is also known that ∠1 is congruent to ∠2. By the Transitive Property of Congruence, ∠ 1 and ∠ 4 are congruent angles. ∠ 1≅ ∠ 2 ∠ 2≅ ∠ 4 ⇒ ∠ 1≅ ∠ 4 Additionally, by the Alternate Interior Angles Theorem, ∠1 is congruent to ∠3. Using the Transitive Property of Congruence one more time, it can be said that ∠ 3 and ∠ 4 are also congruent angles. ∠ 1≅ ∠ 3 ∠ 1≅ ∠ 4 ⇒ ∠ 3≅ ∠ 4 This can be shown in the diagram.

Note that △ACE is divided by l, which is parallel to AE. Therefore, by the Triangle Proportionality Theorem, l divides the other two sides of this triangle proportionally. AD/DC=EB/BC The Converse Isosceles Triangle Theorem states that if two angles in a triangle are congruent, the sides opposite them are congruent. This means that EB is congruent to AB. Therefore, by the definition of congruent segments, they have the same length. AB can be substituted for EB in the above proportion.