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Sagot :
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.
### 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.
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