Given:  △ABC,  BD¯¯bisects  ∠ABC




1​ △ABC​ , BD¯¯¯¯¯ bisects  ∠ABC          Given

2​ ∠AEB≅∠DBC                                         ​Corresponding Angles Theorem

3​ ∠DBC≅∠ABD ​

4∠AEB≅∠ABD∠AEB≅∠ABD                    Transitive Property


6∠AEB≅∠BAE∠AEB≅∠BAE                     Transitive Property

7​AB=EB​                                                     Converse of Isosceles Base Angle Theorem

8​ ADDC=EBBC                                          ​Triangle Proportionality Theorem



POSSIBLE CHOICES FOR 3,5,9            Def of bisector , Substituitin property,  Angle Addition Prostulate,  Alternate inteior angles theorem , Alternave exteerior angles theorem.

Guest Feb 15, 2018

3) Def of Bisector


It is already given info that \(\overline{BD}\) bisects \(\angle ABC\), so you can use the definition of a bisector (which states that a bisector divides a figure into two congruent parts) to make the conclusion.


5) Alternate Interior Angles Theorem


Both \(\angle ABD\) and \(\angle BAE\) have a relationship because the angles are formed on opposite side of the transversal, \(\overline{BA}\), and both lie in the inner region of the parallel segments. This indictates that both angles are alternate interior angles. Since parallel lines and transversals exist in this scenario, it is possible to use the above theorem to conclude the angles' relationship with each other. 


9) Substitution Property (of Equality)


On line 7, it was established that \(AB=EB\). On line 8, it was established that \(\frac{AD}{DC}=\frac{EB}{BC}\). If \(AB=EB\), then we can use the Substitution Property of Equality to conclude that \(\frac{AD}{DC}=\frac{AB}{BC}\), which is identical to the conclusion made. 

TheXSquaredFactor  Feb 15, 2018

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