Question

Please answer 2,3, and 6

Possible answers:
SAS similarity theorem
Alternate interior
Corresponding
AA similarity Postulate
Substitution

Answer 1

The given two column proof is completed as follows;

Statements
`{}` Reasons

1.
`\overline{BE}` ║
`\overline {CD}`
`{}` 1. Given

2. ∠1 ≅ ∠2, ∠3 ≅ ∠4
`{}` 2. Corresponding ∠s Post.

3. ΔABE ~ ΔACD
`{}` 3. AA similarity Postulate

4.
`(AC)/(AB) = (AD)/(AE)`
`{}` 4. Def. of ~ Δs

5. AC = AB + BC; AD = AE + ED
`{}` 5. Segment Add. Post

6.
`(AB + BC)/(AB) = (AE + ED)/(AE)`
`{}` 6. SAS Similarity Theorem

7.
`(AB)/(AB) + (BC)/(AB) = (AE )/(AE) + (AE)/(AE)`
`{}` 7. Distributive property of equality

8.
`1 + (BC)/(AB) = 1 + (AE)/(AE)`
`{}`
`{}` 8.
`Simplify \left((AB)/(AB) =1; (AE )/(AE) =1\right)`

9.
`(BC)/(AB) =(AE)/(AE)`
`{}`
`{}` 9. Subtraction prop. of =

Explanation:

The given two column proof is completed as follows;

Statements
`{}` Reasons

1.
`\overline{BE}` ║
`\overline {CD}`
`{}` 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.
`1 + (BC)/(AB) = 1 + (AE)/(AE)`
`{}`
`{}` 8.
`Simplify \left((AB)/(AB) =1; (AE )/(AE) =1\right)`

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