Converse Triangle Angle Bisector Theorem

Consider △ABC and the segment that connects vertex B with its opposite side. Let D be the point of intersection of the segment from B and AC. Now, CB will be extended to a point E such that BE equals AB. Additionally, a segment from A to E will be constructed.

It is given that BD divides the opposite side in proportion to the sides meeting at B. AD/DC=AB/BC Because BE is equal to AB, by the Substitution Property of Equality BE can be substituted for AB in the proportion. AD/DC=AB/BC substitute AD/DC=BE/BC Therefore, BD is a segment between two sides of △ ACE that divides EC and AC proportionally. Then, by the Converse Triangle Proportionality Theorem it can be stated that EA is parallel to BD.

It is seen that ∠AEB and ∠DBC are corresponding angles. By the Corresponding Angles Theorem, ∠AEB is congruent to ∠DBC. Furthermore, ∠EAB and ∠ABD are alternate interior angles, and by the Alternate Interior Angles Theorem these two angles are also congruent. ∠AEB≅∠CBD ∠EAB≅∠ABD Because BE=AB, by the Isosceles Triangle Theorem ∠AEB is congruent to ∠EAB.

Since ∠CBD and ∠EAB are both congruent to ∠AEB, by the Transitive Property of Congruence, it follows that ∠CBD and ∠EAB are congruent angles. ∠AEB ≅ ∠CBD ∠AEB ≅ ∠EAB ⇓ ∠ CBD ≅ ∠ EAB By the same property, since ∠ABD and ∠CBD are both congruent to ∠EAB, they are congruent angles.

Therefore, by the definition of an angle bisector BD is an angle bisector of the triangle.