Question

the table contains the proof of the theorem of the relationship between slopes of parallel lines. what is the missing statement for step 6? a. slope of \\(\overleftrightarrow{ac}=-\frac{1}{slope of \overleftrightarrow{df}}\\) b. slope of \\(\overleftrightarrow{ac}=\frac{1}{slope of \overleftrightarrow{df}}\\) c. slope of \\(\overleftrightarrow{ac}=- slope of \overleftrightarrow{df}\\) d. slope of \\(\overleftrightarrow{ac}=slope of \overleftrightarrow{df}\\) e. slope o
statement\treason

1. \\(\triangle abc\\) and \\(\triangle def\\) are similar\tgiven
2. \\(\frac{bc}{ef}=\frac{ab}{de}\\)\tproperty of similar triangles
3. \\(\frac{ab}{bc}=\frac{de}{ef}\\)\tproperty of proportion
4. slope of line ac: \\(\frac{bc}{ab}\\)\tdefinition of slope
5. slope of line df: \\(\frac{ef}{de}\\)\tdefinition of slope

6.\tsubstitution property of equality

Explanation:

Step1: Recall slope - definition and similar - triangle properties

We know that for two similar right - triangles formed by two lines, the ratios of their corresponding sides are equal. The slope of a line is defined as the ratio of the vertical change to the horizontal change.
For line \(AC\), slope \(m_{AC}=\frac{BC}{AB}\), and for line \(DF\), slope \(m_{DF}=\frac{EF}{DE}\). Since \(\triangle ABC\sim\triangle DEF\), we have \(\frac{BC}{EF}=\frac{AB}{DE}\), which can be rewritten as \(\frac{BC}{AB}=\frac{EF}{DE}\).

Step2: Use substitution property of equality

By the substitution property of equality, since the slope of line \(AC\) is \(\frac{BC}{AB}\) and the slope of line \(DF\) is \(\frac{EF}{DE}\), and \(\frac{BC}{AB}=\frac{EF}{DE}\), we can say that the slope of line \(AC\) is equal to the slope of line \(DF\).

Answer:

D. slope of \(\overleftrightarrow{AC}\) = slope of \(\overleftrightarrow{DF}\)