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To prove that the distance from the restaurant to the movie theater [tex]\( AC \)[/tex] is the same as the distance from the coffee shop to the estate office [tex]\( DF \)[/tex], we will use a two-column proof format. Here's the step-by-step proof:
[tex]\[ \begin{array}{|c|c|} \hline \text{Statement} & \text{Reason} \\ \hline AB = DE, BC = CD = EF & \text{Given} \\ AB + BC = AC & \text{Segment Addition Postulate} \\ DE + EF = DF & \text{Segment Addition Postulate} \\ AB + BC = DE + EF & \text{Substitution Property of Equality} \\ AC = DF & \text{Transitive Property of Equality} \\ \hline \end{array} \][/tex]
## Detailed Explanation:
1. Given: [tex]\( AB = DE \)[/tex] and [tex]\( BC = CD = EF \)[/tex]
- These are the initial conditions provided in the problem.
2. Segment Addition Postulate: [tex]\( AB + BC = AC \)[/tex]
- According to the Segment Addition Postulate, the sum of the lengths of segments [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex] equals the length of segment [tex]\( AC \)[/tex].
3. Segment Addition Postulate: [tex]\( DE + EF = DF \)[/tex]
- Similarly, according to the Segment Addition Postulate, the sum of the lengths of segments [tex]\( DE \)[/tex] and [tex]\( EF \)[/tex] equals the length of segment [tex]\( DF \)[/tex].
4. Substitution Property of Equality: [tex]\( AB + BC = DE + EF \)[/tex]
- Since [tex]\( AB \)[/tex] is equal to [tex]\( DE \)[/tex] and [tex]\( BC \)[/tex] is equal to [tex]\( EF \)[/tex] (and [tex]\( CD \)[/tex]), we can substitute [tex]\( DE \)[/tex] for [tex]\( AB \)[/tex] and [tex]\( EF \)[/tex] for [tex]\( BC \)[/tex] in the equation from Step 2, giving us [tex]\( DE + EF \)[/tex] as the sum.
5. Transitive Property of Equality: [tex]\( AC = DF \)[/tex]
- By the Transitive Property of Equality, if [tex]\( AB + BC = AC \)[/tex] and [tex]\( DE + EF = DF \)[/tex] are both equal to each other, then [tex]\( AC \)[/tex] must be equal to [tex]\( DF \)[/tex].
Thus, we have proven that the distance from the restaurant to the movie theater [tex]\( AC \)[/tex] is equal to the distance from the coffee shop to the estate office [tex]\( DF \)[/tex].
[tex]\[ \begin{array}{|c|c|} \hline \text{Statement} & \text{Reason} \\ \hline AB = DE, BC = CD = EF & \text{Given} \\ AB + BC = AC & \text{Segment Addition Postulate} \\ DE + EF = DF & \text{Segment Addition Postulate} \\ AB + BC = DE + EF & \text{Substitution Property of Equality} \\ AC = DF & \text{Transitive Property of Equality} \\ \hline \end{array} \][/tex]
## Detailed Explanation:
1. Given: [tex]\( AB = DE \)[/tex] and [tex]\( BC = CD = EF \)[/tex]
- These are the initial conditions provided in the problem.
2. Segment Addition Postulate: [tex]\( AB + BC = AC \)[/tex]
- According to the Segment Addition Postulate, the sum of the lengths of segments [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex] equals the length of segment [tex]\( AC \)[/tex].
3. Segment Addition Postulate: [tex]\( DE + EF = DF \)[/tex]
- Similarly, according to the Segment Addition Postulate, the sum of the lengths of segments [tex]\( DE \)[/tex] and [tex]\( EF \)[/tex] equals the length of segment [tex]\( DF \)[/tex].
4. Substitution Property of Equality: [tex]\( AB + BC = DE + EF \)[/tex]
- Since [tex]\( AB \)[/tex] is equal to [tex]\( DE \)[/tex] and [tex]\( BC \)[/tex] is equal to [tex]\( EF \)[/tex] (and [tex]\( CD \)[/tex]), we can substitute [tex]\( DE \)[/tex] for [tex]\( AB \)[/tex] and [tex]\( EF \)[/tex] for [tex]\( BC \)[/tex] in the equation from Step 2, giving us [tex]\( DE + EF \)[/tex] as the sum.
5. Transitive Property of Equality: [tex]\( AC = DF \)[/tex]
- By the Transitive Property of Equality, if [tex]\( AB + BC = AC \)[/tex] and [tex]\( DE + EF = DF \)[/tex] are both equal to each other, then [tex]\( AC \)[/tex] must be equal to [tex]\( DF \)[/tex].
Thus, we have proven that the distance from the restaurant to the movie theater [tex]\( AC \)[/tex] is equal to the distance from the coffee shop to the estate office [tex]\( DF \)[/tex].
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