Answered

```markdown
4
\begin{tabular}{|c|c|}
\hline
Statements & Reasons \\
\hline
\begin{tabular}{l}
Vertices of [tex]$\triangle ABC$[/tex] are \\
[tex]$A(x_1, y_1)$[/tex], [tex]$B(x_2, y_2)$[/tex], and [tex]$C(x_3, y_3)$[/tex].
\end{tabular} & Given \\
\hline
\begin{tabular}{l}
Use rigid transformations to transform \\
[tex]$\triangle ABC$[/tex] into [tex]$\triangle A'B'C'$[/tex] such that \\
vertex [tex]$A'$[/tex] is at the origin and [tex]$\overline{A'C'}$[/tex] \\
lies on the [tex]$x$[/tex]-axis in the positive direction.
\end{tabular} & \begin{tabular}{l}
In the coordinate plane, any point can \\
be moved to any other point using rigid \\
transformations and any line can be \\
moved to any other line using rigid \\
transformations.
\end{tabular} \\
\hline
\begin{tabular}{l}
Any property that is true for [tex]$\triangle A'B'C'$[/tex] will \\
be true for [tex]$\triangle ABC$[/tex].
\end{tabular} & Definition of congruence \\
\hline
\begin{tabular}{l}
Let [tex]$r$[/tex], [tex]$s$[/tex], and [tex]$t$[/tex] be real numbers such that \\
vertices of [tex]$\triangle A'B'C'$[/tex] are [tex]$A'(0, 0)$[/tex], \\
[tex]$B'(r, s)$[/tex], and [tex]$C'(2t, 0)$[/tex].
\end{tabular} & Defining constants \\
\hline
\begin{tabular}{l}
Let [tex]$D'$[/tex], [tex]$E'$[/tex], and [tex]$F'$[/tex] be the midpoints of \\
[tex]$\overline{B'C'}$[/tex], [tex]$\overline{A'C'}$[/tex], and [tex]$\overline{A'B'}$[/tex] respectively.
\end{tabular} & Defining points \\
\hline
\begin{tabular}{l}
[tex]$D'=(r, s)$[/tex] \\
[tex]$E'=(r+t, s)$[/tex] \\
[tex]$F'=(t, 0)$[/tex]
\end{tabular} & Definition of midpoints \\
\hline
\end{tabular}

What is step 10 in this proof?

A. Statement: line [tex]$AE'$[/tex] and line [tex]$BF'$[/tex] are concurrent.
Reason: Definition of concurrence

B. Statement: All three lines share point [tex]$P$[/tex].
Reason: Definition of midpoint

C. Statement: Point [tex]$P$[/tex] lies on line [tex]$AE'$[/tex] and line [tex]$BF'$[/tex].
Reason: Algebra

D. Statement: Point [tex]$P$[/tex] lies on line [tex]$CD$[/tex].
Reason: The coordinates of [tex]$P$[/tex] satisfy the equation of [tex]$CD$[/tex].

E. Statement: Two of the three medians share point [tex]$P$[/tex].
Reason: Using point-slope formula
```



Answer :

GinnyAnswer
Let's solve this problem step-by-step using the given solution:

### Given Information:
1. Vertices of triangle [tex]\( \triangle ABC \)[/tex]:
- [tex]\( A(x_1, y_1) \)[/tex]
- [tex]\( B(x_2, y_2) \)[/tex]
- [tex]\( C(x_3, y_3) \)[/tex]

2. Transformations Applied:
- Triangle [tex]\( \triangle ABC \)[/tex] is transformed into [tex]\( \triangle A'B'C' \)[/tex] using rigid transformations such that:
- [tex]\( A'(0,0) \)[/tex] (vertex [tex]\( A' \)[/tex] is at the origin)
- [tex]\( \overline{A'C'} \)[/tex] lies on the x-axis in the positive direction

3. Congruence:
- Any property true for [tex]\( \triangle A'B'C' \)[/tex] is true for [tex]\( \triangle ABC \)[/tex] because they are congruent by transformation.

4. Vertices of [tex]\( \triangle A'B'C' \)[/tex]:
- [tex]\( A'(0,0) \)[/tex]
- [tex]\( B'(r, 2s) \)[/tex]
- [tex]\( C'(2t, 0) \)[/tex]
- Where [tex]\( r \)[/tex], [tex]\( s \)[/tex], and [tex]\( t \)[/tex] are real numbers.

5. Midpoints:
- [tex]\( D' \)[/tex], [tex]\( E' \)[/tex], and [tex]\( F' \)[/tex] are midpoints of [tex]\( \overline{B'C'} \)[/tex], [tex]\( \overline{A'C'} \)[/tex], and [tex]\( \overline{A'B'} \)[/tex] respectively:
- [tex]\( D'(r, s) \)[/tex]
- [tex]\( E'(r+t, s) \)[/tex]
- [tex]\( F'(t, 0) \)[/tex]

### Analysis:
To answer what step 10 in this proof involves, let's focus on what each option implies:
- Option A: "Line [tex]\( AE' \)[/tex] and line [tex]\( BF \)[/tex] are concurrent" refers to the property of concurrence, meaning these lines meet at a common point.
- Option B: "All three lines share point [tex]\( P \)[/tex]" would include all three medians, but we specifically need to show concurrence of at least two at this step.
- Option C: "Point [tex]\( P \)[/tex] lies on line [tex]\( AE \)[/tex] and line [tex]\( BF' \)[/tex]" involves algebraic reasoning to show the point of intersection.
- Option D: "Point [tex]\( P \)[/tex] lies on line [tex]\( CD \)[/tex]" involves checking if the point [tex]\( P \)[/tex] satisfies the equation of line [tex]\( CD \)[/tex].
- Option E: "Two of the three medians share point [tex]\( P \)[/tex]" involves using point-slope formula to show that two medians intersect at point [tex]\( P \)[/tex].

### Correct Step:
Given the information and context, the step in the proof should logically align with showing concurrence:

Step 10 Statement:
- Statement: Line [tex]\( AE' \)[/tex] and line [tex]\( BF \)[/tex] are concurrent.
- Reason: Definition of concurrence.

Therefore, the correct answer is:
A. Statement: Line [tex]\( AE' \)[/tex] and line [tex]\( BF \)[/tex] are concurrent. Reason: Definition of concurrence.