Please answer 2,3, and 6
Possible answers:
SAS similarity theorem
Alternate interior
Corresponding
AA similarity Postulate
Substitution

Please answer 23 and 6 Possible answers SAS similarity theorem Alternate interior Corresponding AA similarity Postulate Substitution class=

Respuesta :

Answer:

The given two column proof is completed as follows;

Statements    [tex]{}[/tex]                                               Reasons

1. [tex]\overline{BE}[/tex] ║ [tex]\overline {CD}[/tex]    [tex]{}[/tex]                                               1. Given

2. ∠1 ≅ ∠2, ∠3 ≅ ∠4 [tex]{}[/tex]                                    2. Corresponding ∠s Post.

3. ΔABE ~ ΔACD [tex]{}[/tex]                                          3. AA similarity Postulate

4. [tex]\dfrac{AC}{AB} = \dfrac{AD}{AE}[/tex]    [tex]{}[/tex]                                              4. Def. of ~ Δs

5. AC = AB + BC; AD = AE + ED   [tex]{}[/tex]                5. Segment Add. Post

6. [tex]\dfrac{AB + BC}{AB} = \dfrac{AE + ED}{AE}[/tex]   [tex]{}[/tex]                           6. SAS Similarity Theorem

7. [tex]\dfrac{AB}{AB} + \dfrac{BC}{AB} = \dfrac{AE }{AE} + \dfrac{AE}{AE}[/tex]   [tex]{}[/tex]                          7. Distributive property of equality

8. [tex]1 + \dfrac{BC}{AB} = 1 + \dfrac{AE}{AE}[/tex]  [tex]{}[/tex]    [tex]{}[/tex]                               8. [tex]Simplify \left(\dfrac{AB}{AB} =1; \dfrac{AE }{AE} =1\right)[/tex]

9. [tex]\dfrac{BC}{AB} =\dfrac{AE}{AE}[/tex]          [tex]{}[/tex]    [tex]{}[/tex]                                    9. Subtraction prop. of =

Step-by-step explanation:

The given two column proof is completed as follows;

Statements    [tex]{}[/tex]                                               Reasons

1. [tex]\overline{BE}[/tex] ║ [tex]\overline {CD}[/tex]    [tex]{}[/tex]                                               1. Given

2. Corresponding angles are congruent (postulate)

3. Angle Angle, AA similarity Postulate

4. Def. of ~ Δs Definition of similar triangles

5. Segment Addition Postulate

6. Side-Angle-Side SAS Similarity Theorem

7. Distributive property of equality

8. [tex]1 + \dfrac{BC}{AB} = 1 + \dfrac{AE}{AE}[/tex]  [tex]{}[/tex]    [tex]{}[/tex]                               8. [tex]Simplify \left(\dfrac{AB}{AB} =1; \dfrac{AE }{AE} =1\right)[/tex]

9. Subtraction property of equality, by subtracting 1 from both sides

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