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zvonat [6]
2 years ago
13

Please answer 2,3, and 6

Mathematics
1 answer:
aleksandrvk [35]2 years ago
8 0

Answer:

The given two column proof is completed as follows;

Statements    {}                                               Reasons

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

2. ∠1 ≅ ∠2, ∠3 ≅ ∠4 {}                                    2. <u>Corresponding</u> ∠s Post.

3. ΔABE ~ ΔACD {}                                          3. <u>AA similarity Postulate</u>

4. \dfrac{AC}{AB} = \dfrac{AD}{AE}    {}                                              4. Def. of ~ Δs

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

6. \dfrac{AB + BC}{AB} = \dfrac{AE + ED}{AE}   {}                           6. <u>SAS Similarity Theorem</u>

7. \dfrac{AB}{AB} + \dfrac{BC}{AB}  = \dfrac{AE }{AE} + \dfrac{AE}{AE}   {}                          7. Distributive property of equality

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

9. \dfrac{BC}{AB}  =\dfrac{AE}{AE}          {}    {}                                    9. Subtraction prop. of =

Step-by-step 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 + \dfrac{BC}{AB}  = 1 + \dfrac{AE}{AE}  {}    {}                               8. Simplify \left(\dfrac{AB}{AB} =1;  \dfrac{AE }{AE} =1\right)

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

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