Question

select the correct answer from each drop - down menu.
in the figure, ( cd = ef ) and ( ab = ce ). complete the statements to prove that ( ab = df ).
( cd + de = ef + de ) by the (\boldsymbol{\text{drop - down menu}}) property of equality.
( ce = cd + de ) and ( df = ef + de ) by (\boldsymbol{\text{drop - down menu}})
( ce = df ) by the (\boldsymbol{\text{drop - down menu}}) property of equality.
given, ( ab = ce ) and ( ce = df ) implies ( ab = df ) by the (\boldsymbol{\text{drop - down menu}}) property of equality.

Explanation:

1. For the first blank: Adding the same value \(DE\) to both sides of the equality \(CD=EF\) uses the Addition Property of Equality, which states that if \(a=b\), then \(a+c=b+c\).
2. For the second blank: The statements \(CE = CD + DE\) and \(DF = EF + DE\) follow from the Segment Addition Postulate, which says that if a point lies between two endpoints of a segment, the total length is the sum of the smaller segments.
3. For the third blank: Substituting \(CD+DE=CE\) and \(EF+DE=DF\) (from the second step) into \(CD+DE=EF+DE\) gives \(CE=DF\), which uses the Substitution Property of Equality (if \(a=b\) and \(a=c\), \(b=c\)).
4. For the fourth blank: Given \(AB=CE\) and \(CE=DF\), concluding \(AB=DF\) uses the Transitive Property of Equality, which states that if \(a=b\) and \(b=c\), then \(a=c\).

Answer:

1. Addition
2. Segment Addition Postulate
3. Substitution
4. Transitive