Question

segment proofs
directions: complete the proof to the right using the statements and reasons.
given: c is the midpoint of \\(\overline{ae}\\);
b is the midpoint of \\(\overline{ac}\\);
d is the midpoint of \\(\overline{ce}\\)
prove: \\(\overline{bc} \cong \overline{cd}\\)
statements

1. c is the midpoint of \\(\overline{ae}\\);

b is the midpoint of \\(\overline{ac}\\);
d is the midpoint of \\(\overline{ce}\\)

1. \\(ac = ce\\); \\(ab = bc\\); \\(cd = de\\)
2. \\(ab + bc = ac\\); \\(cd + de = ce\\)
3. \\(bc + bc = cd + cd\\)
4. \\(2bc = 2cd\\)
5. \\(bc = cd\\)
6. \\(\overline{bc} \cong \overline{cd}\\)
7. place statement here
8. place statement here

reasons

1. given
2. definition of midpoint
3. segment addition postulate
4. substitution property
5. simplify
6. division property
7. definition of congruence
8. place reason here
9. place reason here

\\(ab + bc = cd + de\\)
\\(cd + de = ac\\)

Explanation:

To complete the proof, we analyze the remaining steps (steps 8 and 9) by using the transitive property and substitution:

Step 8:

Use transitive property with \( AC = CE \) (from step 2) and \( CD + DE = CE \) (from step 3), and \( AB + BC = AC \) (from step 3) and \( AB + BC = CD + DE \) (from the left box).
Statement: \( AC = CD + DE \)
Reason: Transitive Property (since \( AC = CE \) and \( CD + DE = CE \), so \( AC = CD + DE \))

Step 9:

Substitute \( AB + BC \) for \( AC \) (from step 3: \( AB + BC = AC \)) into \( AC = CD + DE \).
Statement: \( AB + BC = CD + DE \)
Reason: Substitution Property (replace \( AC \) with \( AB + BC \) in \( AC = CD + DE \))

Final Answers for Steps 8 and 9:

1. Statement: \( AC = CD + DE \); Reason: Transitive Property
2. Statement: \( AB + BC = CD + DE \); Reason: Substitution Property

(Note: The left box provides \( AB + BC = CD + DE \) as a hint, confirming the statement for step 9.)

Answer:

Substitute \( AB + BC \) for \( AC \) (from step 3: \( AB + BC = AC \)) into \( AC = CD + DE \).
Statement: \( AB + BC = CD + DE \)
Reason: Substitution Property (replace \( AC \) with \( AB + BC \) in \( AC = CD + DE \))

Final Answers for Steps 8 and 9:

1. Statement: \( AC = CD + DE \); Reason: Transitive Property
2. Statement: \( AB + BC = CD + DE \); Reason: Substitution Property

(Note: The left box provides \( AB + BC = CD + DE \) as a hint, confirming the statement for step 9.)