Given: $\overline{BE}$bisects $\angle DBA;\angle 3\cong \angle 1$

Prove: $\overline{CD}||\overline{BE}$

The theorem 3-5 can be used which states that the relation between two lines is parallel only when a separate line travels through these two lines and the angles which are positioned on the alternate interior side represents the congruent angles.

The definition of transitive property of congruence states that whenever two angles are congruent and if any one of them is congruent to another angle that is beside those two angles then the second angle from first two angles also congruent to the another angle. According

According to the provided bisection of angle $m\angle DBE=m\angle EBA$which means that $m\angle 2=m\angle 3$. As the measurements are equal so this will result in $\angle 2\cong \angle 3$.

By using the transitive property, $\angle 2\cong \angle 3$and provided angles the angle $\angle 1\cong \angle 2$.

The angles $\angle 1$and $\angle 2$are associated with the lines $\overline{CD}$and $\overline{BE}$because a separate line role="math" localid="1637916925571" $\overline{DB}$is travelling through these lines and result in formation of angles $\angle 1$and $\angle 2$. According to the theorem 3-5 the lines $\overline{CD}$and $\overline{BE}$are having a parallel connection between them.