Given: AC bisects \angle BAD

Question

Given: AC bisects \angle BAD
\angle ABC=\angle ADC
Write a two-column proof to prove= BC=DC

Answer 1

AC bisects ∠BAD, => ∠BAC=∠CAD ..... (1)

thus in ΔABC and ΔADC,    ∠ABC=∠ADC (given), 

                                       ∠BAC=∠CAD      [from (1)],

AC (opposite side side of ∠ABC) = AC (opposite side side of ∠ADC), the common side between ΔABC and ΔADC

Hence,      by AAS axiom, ΔABC ≅ ΔADC,

Therefore,  BC (opposite side side of ∠BAC) = DC (opposite side side of ∠CAD),  since (1)

Hence, BC=DC proved.

Answer 2

Dear Emma,

     This is what you need to do to construct the proof.  The bisection and the two segments that they want you to prove that they are equal make two triangles.  You need to first prove that these triangles, Triangle ABC and Triangle ADC are equal by Angle-Angle-side.  Then  you can say that the two segments, BC and DC, are equal by "corresponding parts of congruent triangles are congruent, (CPCTC).  By the way, both these triangles are sharing one side, AC. You should draw a picture before you begin the proof.

 

Given:   Segment AC bisects ∠BAD, ∠Abc =∠ADC                  Prove:  Segment BC=Segment DC

 

___Statements___________________________Reasons_________________________

 

1.  Seg. AC bisects ∠BAD                                 1.  Given

 

2.  (m=measure) m∠BAC=∠CAD                      2.  Definition of angle bisector

 

    Therefore, m∠BAC≅∠CAD.

 

3.  AC=AC                                                      3.  reflexive property

 

4.  ∠ABC =∠ADC                                            4.  Given

 

5.  Triangle ABC≅Triangle ADC                        5.  Angle-Angle-Side (AAS postulate).

 

6.  Therefore, BC=DC                                     6. CPCTC

 

 

Note:  (Corresponding parts of congruent triangles are congruent.)

 

        A-A-S= Two pairs of angles correspond, but there is a shared side that is not between the angles.

 

 

I hope that I have helped you.  If I have, please give me a thumbs up.

                

                                                                     Susan C.